718 lines
22 KiB
C
718 lines
22 KiB
C
/* $NetBSD: rb.c,v 1.7 2006/09/06 20:01:57 thorpej Exp $ */
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/*-
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* Copyright (c) 2001 The NetBSD Foundation, Inc.
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* All rights reserved.
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*
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* This code is derived from software contributed to The NetBSD Foundation
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* by Matt Thomas <matt@3am-software.com>.
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*
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* Redistribution and use in source and binary forms, with or without
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* modification, are permitted provided that the following conditions
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* are met:
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* 1. Redistributions of source code must retain the above copyright
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* notice, this list of conditions and the following disclaimer.
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* 2. Redistributions in binary form must reproduce the above copyright
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* notice, this list of conditions and the following disclaimer in the
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* documentation and/or other materials provided with the distribution.
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* 3. All advertising materials mentioning features or use of this software
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* must display the following acknowledgement:
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* This product includes software developed by the NetBSD
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* Foundation, Inc. and its contributors.
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* 4. Neither the name of The NetBSD Foundation nor the names of its
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* contributors may be used to endorse or promote products derived
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* from this software without specific prior written permission.
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*
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* THIS SOFTWARE IS PROVIDED BY THE NETBSD FOUNDATION, INC. AND CONTRIBUTORS
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* ``AS IS'' AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED
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* TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR
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* PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE FOUNDATION OR CONTRIBUTORS
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* BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR
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* CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF
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* SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS
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* INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN
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* CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE)
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* ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE
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* POSSIBILITY OF SUCH DAMAGE.
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*/
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#if !defined(_KERNEL) && !defined(_STANDALONE)
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#include <sys/types.h>
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#include <stddef.h>
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#include <assert.h>
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#include <stdbool.h>
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#define KASSERT(s) assert(s)
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#else
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#include <lib/libkern/libkern.h>
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#endif
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#include "rb.h"
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static void rb_tree_swap_nodes(struct rb_tree *, struct rb_node *, int);
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static void rb_tree_insert_rebalance(struct rb_tree *, struct rb_node *);
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static void rb_tree_removal_rebalance(struct rb_tree *, struct rb_node *,
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unsigned int);
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#ifndef NDEBUG
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static bool rb_tree_check_node(const struct rb_tree *, const struct rb_node *,
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const struct rb_node *, bool);
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#endif
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/*
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* Rather than testing for the NULL everywhere, all terminal leaves are
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* pointed to this node. Note that by setting it to be const, that on
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* some architectures trying to write to it will cause a fault.
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*/
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static const struct rb_node sentinel_node = { .rb_sentinel = 1 };
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void
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rb_tree_init(struct rb_tree *rbt, rb_compare_nodes_fn compare_nodes,
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rb_compare_key_fn compare_key, rb_print_node_fn print_node)
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{
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TAILQ_INIT(&rbt->rbt_nodes);
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rbt->rbt_count = 0;
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rbt->rbt_compare_nodes = compare_nodes;
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rbt->rbt_compare_key = compare_key;
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rbt->rbt_print_node = print_node;
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*((const struct rb_node **)&rbt->rbt_root) = &sentinel_node;
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}
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/*
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* Swap the location and colors of 'self' and its child @ which. The child
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* can not be a sentinel node.
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*/
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static void
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rb_tree_swap_nodes(struct rb_tree *rbt, struct rb_node *old_father, int which)
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{
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const unsigned int other = which ^ RB_OTHER;
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struct rb_node * const grandpa = old_father->rb_parent;
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struct rb_node * const old_child = old_father->rb_nodes[which];
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struct rb_node * const new_father = old_child;
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struct rb_node * const new_child = old_father;
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unsigned int properties;
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KASSERT(!RB_SENTINEL_P(old_child));
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KASSERT(old_child->rb_parent == old_father);
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KASSERT(rb_tree_check_node(rbt, old_father, NULL, false));
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KASSERT(rb_tree_check_node(rbt, old_child, NULL, false));
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KASSERT(RB_ROOT_P(old_father) || rb_tree_check_node(rbt, grandpa, NULL, false));
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/*
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* Exchange descendant linkages.
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*/
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grandpa->rb_nodes[old_father->rb_position] = new_father;
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new_child->rb_nodes[which] = old_child->rb_nodes[other];
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new_father->rb_nodes[other] = new_child;
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/*
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* Update ancestor linkages
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*/
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new_father->rb_parent = grandpa;
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new_child->rb_parent = new_father;
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/*
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* Exchange properties between new_father and new_child. The only
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* change is that new_child's position is now on the other side.
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*/
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properties = old_child->rb_properties;
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new_father->rb_properties = old_father->rb_properties;
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new_child->rb_properties = properties;
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new_child->rb_position = other;
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/*
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* Make sure to reparent the new child to ourself.
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*/
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if (!RB_SENTINEL_P(new_child->rb_nodes[which])) {
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new_child->rb_nodes[which]->rb_parent = new_child;
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new_child->rb_nodes[which]->rb_position = which;
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}
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KASSERT(rb_tree_check_node(rbt, new_father, NULL, false));
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KASSERT(rb_tree_check_node(rbt, new_child, NULL, false));
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KASSERT(RB_ROOT_P(new_father) || rb_tree_check_node(rbt, grandpa, NULL, false));
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}
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void
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rb_tree_insert_node(struct rb_tree *rbt, struct rb_node *self)
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{
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struct rb_node *prev, *next, *parent, *tmp;
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unsigned int position;
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self->rb_properties = 0;
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prev = NULL;
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next = NULL;
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tmp = rbt->rbt_root;
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parent = (struct rb_node *)&rbt->rbt_root;
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position = RB_LEFT;
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/*
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* Find out where to place this new leaf.
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*/
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while (!RB_SENTINEL_P(tmp)) {
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const int diff = (*rbt->rbt_compare_nodes)(tmp, self);
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parent = tmp;
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KASSERT(diff != 0);
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if (diff < 0) {
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position = RB_LEFT;
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next = parent->rb_left;
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prev = NULL;
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} else {
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position = RB_RIGHT;
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prev = parent->rb_right;
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next = NULL;
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}
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tmp = parent->rb_nodes[position];
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}
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/*
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* Verify our sequential position
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*/
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if (prev != NULL && !RB_SENTINEL_P(prev) && next == NULL)
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next = TAILQ_NEXT(prev, rb_link);
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if (prev == NULL && (next != NULL && !RB_SENTINEL_P(next)))
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next = TAILQ_PREV(next, rb_node_qh, rb_link);
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KASSERT(prev == NULL || RB_SENTINEL_P(prev) || (*rbt->rbt_compare_nodes)(prev, self) > 0);
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KASSERT(next == NULL || RB_SENTINEL_P(next) || (*rbt->rbt_compare_nodes)(self, next) < 0);
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/*
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* Initialize the node and insert as a leaf into the tree.
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*/
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self->rb_parent = parent;
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self->rb_position = position;
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if (__predict_false(parent == (struct rb_node *) &rbt->rbt_root)) {
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RB_MARK_ROOT(self);
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} else {
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KASSERT(position == RB_LEFT || position == RB_RIGHT);
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KASSERT(!RB_ROOT_P(self)); /* Already done */
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}
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KASSERT(RB_SENTINEL_P(parent->rb_nodes[position]));
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self->rb_left = parent->rb_nodes[position];
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self->rb_right = parent->rb_nodes[position];
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parent->rb_nodes[position] = self;
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KASSERT(self->rb_left == &sentinel_node &&
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self->rb_right == &sentinel_node);
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/*
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* Insert the new node into a sorted list for easy sequential access
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*/
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rbt->rbt_count++;
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if (next != NULL && !RB_SENTINEL_P(next)) {
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TAILQ_INSERT_BEFORE(next, self, rb_link);
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} else {
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TAILQ_INSERT_TAIL(&rbt->rbt_nodes, self, rb_link);
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}
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#if 0
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/*
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* Validate the tree before we rebalance
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*/
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rb_tree_check(rbt, false);
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#endif
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/*
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* Rebalance tree after insertion
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*/
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rb_tree_insert_rebalance(rbt, self);
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#if 0
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/*
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* Validate the tree after we rebalanced
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*/
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rb_tree_check(rbt, true);
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#endif
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}
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static void
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rb_tree_insert_rebalance(struct rb_tree *rbt, struct rb_node *self)
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{
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RB_MARK_RED(self);
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while (!RB_ROOT_P(self) && RB_RED_P(self->rb_parent)) {
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const unsigned int which =
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(self->rb_parent == self->rb_parent->rb_parent->rb_left
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? RB_LEFT
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: RB_RIGHT);
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const unsigned int other = which ^ RB_OTHER;
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struct rb_node * father = self->rb_parent;
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struct rb_node * grandpa = father->rb_parent;
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struct rb_node * const uncle = grandpa->rb_nodes[other];
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KASSERT(!RB_SENTINEL_P(self));
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/*
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* We are red and our parent is red, therefore we must have a
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* grandfather and he must be black.
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*/
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KASSERT(RB_RED_P(self)
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&& RB_RED_P(father)
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&& RB_BLACK_P(grandpa));
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if (RB_RED_P(uncle)) {
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/*
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* Case 1: our uncle is red
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* Simply invert the colors of our parent and
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* uncle and make our grandparent red. And
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* then solve the problem up at his level.
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*/
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RB_MARK_BLACK(uncle);
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RB_MARK_BLACK(father);
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RB_MARK_RED(grandpa);
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self = grandpa;
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continue;
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}
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/*
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* Case 2&3: our uncle is black.
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*/
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if (self == father->rb_nodes[other]) {
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/*
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* Case 2: we are on the same side as our uncle
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* Swap ourselves with our parent so this case
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* becomes case 3. Basically our parent becomes our
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* child.
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*/
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self = father;
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rb_tree_swap_nodes(rbt, self, other);
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father = self->rb_parent;
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grandpa = father->rb_parent;
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}
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KASSERT(RB_RED_P(self) && RB_RED_P(father));
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KASSERT(grandpa->rb_nodes[which] == father);
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/*
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* Case 3: we are opposite a child of a black uncle.
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* Swap our parent and grandparent. Since our grandfather
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* is black, our father will become black and our new sibling
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* (former grandparent) will become red.
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*/
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rb_tree_swap_nodes(rbt, grandpa, which);
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KASSERT(RB_RED_P(self) && RB_BLACK_P(father));
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break;
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}
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/*
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* Final step: Set the root to black.
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*/
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RB_MARK_BLACK(rbt->rbt_root);
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}
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struct rb_node *
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rb_tree_find(struct rb_tree *rbt, void *key)
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{
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struct rb_node *parent = rbt->rbt_root;
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while (!RB_SENTINEL_P(parent)) {
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const int diff = (*rbt->rbt_compare_key)(parent, key);
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if (diff == 0)
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return parent;
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parent = parent->rb_nodes[diff > 0];
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}
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return NULL;
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}
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/*
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*
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*/
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void
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rb_tree_remove_node(struct rb_tree *rbt, struct rb_node *self)
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{
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/*
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* Easy case, there are no children. This means this node is
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* at rank [N] (the bottom) of the red-black tree. We can simple just
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* prune it. This leave the problem of identifying the node removed.
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* Since the node "left" is a sentinel it doesn't carry any information.
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* Instead we identify the removed node by parent and position.
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*/
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if (RB_CHILDLESS_P(self)) {
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KASSERT(rb_tree_check_node(rbt, self, NULL, false));
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self->rb_parent->rb_nodes[self->rb_position] = self->rb_left;
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if (RB_BLACK_P(self) && !RB_ROOT_P(self))
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rb_tree_removal_rebalance(rbt,
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self->rb_parent, self->rb_position);
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KASSERT(RB_ROOT_P(self) || rb_tree_check_node(rbt, self->rb_parent, NULL, true));
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goto done;
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}
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/*
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* If this node only has one valid child, that child must be itself
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* childless. This is because a red-black tree only has leaves at
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* two ranks, [N-1] and [N]. If this node only has one child at this
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* rank, this must be rank [N-1] since all nodes at rank [N] are
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* childless. Thus our child must be at rank [N] and therevore must
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* be childless.
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*
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* Promoting this node's child to rank [N-1] and giving this node's
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* color allows us to pretend we really removed a node of rank [N].
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* Now if the child was red, even better since we can just return
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* knowing that removing a red node from a red-black tree still
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* leaves a valid red-black tree.
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*
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* D \ D \ D
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* B F > B E > B E
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* A C E _ / A C _ F / A C _ x
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*
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* We are removing F, but F has a child E. So we swap E and F (in
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* theory but we really replace F by E thereby removing F from the
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* tree).
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*/
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if (!RB_TWOCHILDREN_P(self)) {
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const unsigned int which = RB_LEFT_SENTINEL_P(self) ? RB_RIGHT : RB_LEFT;
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const unsigned int other = which ^ RB_OTHER;
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struct rb_node * const new_self = self->rb_nodes[which];
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const bool was_black = RB_BLACK_P(new_self);
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KASSERT(RB_CHILDLESS_P(new_self));
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KASSERT(rb_tree_check_node(rbt, new_self, NULL, false));
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/*
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* Copy self to new_self.
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*/
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self->rb_parent->rb_nodes[self->rb_position] = new_self;
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new_self->rb_parent = self->rb_parent;
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new_self->rb_properties = self->rb_properties;
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new_self->rb_nodes[which] = new_self->rb_nodes[other];
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KASSERT(rb_tree_check_node(rbt, new_self, NULL, false));
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if (was_black) {
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/*
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* Now rebalance
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*/
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rb_tree_removal_rebalance(rbt, new_self, other);
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} else {
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/*
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* If new_self was red, then self must have been black.
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*/
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KASSERT(RB_BLACK_P(self));
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/* RB_MARK_BLACK(new_self); */
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}
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KASSERT(rb_tree_check_node(rbt, new_self, NULL, true));
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goto done;
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}
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/*
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* The node to be removed is in the interior of the red-black tree.
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* This results in a difficulty in this red-black implementation
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* since contrary to a traditional implemenation the interior nodes
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* contain the keys while the leaves have no keys.
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*
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* D | C
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* B F | B F
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* A C E G | A x E G
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* 1 2 3 4 5 6 7 | 1 2 3 5 6 7
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*
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* We need to remove D. Whom do we swap with, C or E? It doesn't
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* matter since they are both at rank [N]. We put C as root and
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* then pretend D was a child B. Even though D's key would never
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* have been valid at C, since it's now deleted it's key is
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* irrevalent. We simply claim to have deleted right(B).
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*
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* D | E
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* B F | B F
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* A E G | A x G
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* 1 2 4 5 6 7 | 1 2 5 6 7
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*
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* We need to remove D. Whom do we swap with? E, of course. E is
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* the closest node of rank [N]. Always choose the highest rank you
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* can. We make E root and then pretend D was a child F. Again D's
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* key is irrevalent. We simply claim to have deleted left(F).
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*
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* D | D
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* B F | B G
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* A G | A F
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* 1 2 4 6 7 | 1 2 4 6 7
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*
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* We need to remove D. Whom do we swap with? B and F are equally
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* bad since they both have one child. So we swap one and that
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* reduces to the previous case.
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*/
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if (RB_TWOCHILDREN_P(self)) {
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struct rb_node * const next = TAILQ_NEXT(self, rb_link);
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struct rb_node * const prev = TAILQ_PREV(self, rb_node_qh, rb_link);
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struct rb_node * new_self;
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struct rb_node * parent;
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unsigned int which;
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bool was_black;
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KASSERT(next != NULL && prev != NULL);
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KASSERT(!RB_TWOCHILDREN_P(next));
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KASSERT(!RB_TWOCHILDREN_P(prev));
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/*
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* First, pick the childless victim.
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* Now we two possible victims
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* Otherwise, if root, pick from the side with more nodes.
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* Lastly, pick from the side closest to the root.
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*/
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if (RB_CHILDLESS_P(next) && !RB_CHILDLESS_P(prev)) {
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new_self = next;
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} else if (RB_CHILDLESS_P(prev) && !RB_CHILDLESS_P(next)) {
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new_self = prev;
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} else if (RB_ROOT_P(self)) {
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new_self = prev;
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} else {
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new_self = next;
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}
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KASSERT(new_self != rbt->rbt_root);
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KASSERT(!RB_TWOCHILDREN_P(new_self));
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if (RB_CHILDLESS_P(new_self)) {
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KASSERT(rb_tree_check_node(rbt, new_self, NULL, false));
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which = new_self->rb_position;
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parent = new_self->rb_parent;
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was_black = RB_BLACK_P(new_self);
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parent->rb_nodes[which] = new_self->rb_left;
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KASSERT(rb_tree_check_node(rbt, parent, NULL, false));
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} else {
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/*
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* New self has one child. And we know it's on the
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* opposite side than what we want. So we want to
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* replace new_self with it and then self with new_self.
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*/
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which = new_self->rb_position;
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parent = new_self->rb_nodes[which ^ RB_OTHER];
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was_black = RB_BLACK_P(parent);
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KASSERT(RB_SENTINEL_P(new_self->rb_nodes[which]));
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KASSERT(!RB_SENTINEL_P(parent));
|
||
KASSERT(RB_CHILDLESS_P(parent));
|
||
|
||
/*
|
||
* Move parent to new_self's location in the tree.
|
||
*/
|
||
new_self->rb_parent->rb_nodes[new_self->rb_position] = parent;
|
||
parent->rb_parent = new_self->rb_parent;
|
||
parent->rb_properties = new_self->rb_properties;
|
||
KASSERT(rb_tree_check_node(rbt, parent, NULL, false));
|
||
}
|
||
|
||
/*
|
||
* Move new_self to our location in the tree.
|
||
*/
|
||
self->rb_parent->rb_nodes[self->rb_position] = new_self;
|
||
new_self->rb_left = self->rb_left;
|
||
new_self->rb_right = self->rb_right;
|
||
new_self->rb_parent = self->rb_parent;
|
||
new_self->rb_properties = self->rb_properties;
|
||
new_self->rb_left->rb_parent = new_self;
|
||
new_self->rb_right->rb_parent = new_self;
|
||
KASSERT(rb_tree_check_node(rbt, new_self, NULL, false));
|
||
|
||
/*
|
||
* Rebalance if the node was
|
||
*/
|
||
if (was_black)
|
||
rb_tree_removal_rebalance(rbt, parent, which);
|
||
}
|
||
|
||
done:
|
||
/*
|
||
* Remove ourselves from the node list and decrement the count.
|
||
*/
|
||
TAILQ_REMOVE(&rbt->rbt_nodes, self, rb_link);
|
||
rbt->rbt_count--;
|
||
}
|
||
|
||
static void
|
||
rb_tree_removal_rebalance(struct rb_tree *rbt, struct rb_node *parent,
|
||
unsigned int which)
|
||
{
|
||
KASSERT(!RB_SENTINEL_P(parent));
|
||
KASSERT(RB_SENTINEL_P(parent->rb_nodes[which]));
|
||
|
||
while (RB_BLACK_P(parent->rb_nodes[which])) {
|
||
unsigned int other = which ^ RB_OTHER;
|
||
struct rb_node *brother = parent->rb_nodes[other];
|
||
|
||
KASSERT(!RB_SENTINEL_P(brother));
|
||
/*
|
||
* For cases 1, 2a, and 2b, our brother's children must
|
||
* be black.
|
||
*/
|
||
if (RB_BLACK_P(brother->rb_left)
|
||
&& RB_BLACK_P(brother->rb_right)) {
|
||
/*
|
||
* Case 1: Our brother is red, swap its position
|
||
* (and colors) with our parent. This is now case 2b.
|
||
*
|
||
* B -> D
|
||
* x d -> b E
|
||
* C E -> x C
|
||
*/
|
||
if (RB_RED_P(brother)) {
|
||
KASSERT(RB_BLACK_P(parent));
|
||
rb_tree_swap_nodes(rbt, parent, other);
|
||
brother = parent->rb_nodes[other];
|
||
KASSERT(RB_BLACK_P(parent->rb_nodes[which]));
|
||
KASSERT(!RB_SENTINEL_P(brother));
|
||
KASSERT(RB_RED_P(parent));
|
||
KASSERT(RB_BLACK_P(brother->rb_left)
|
||
&& RB_BLACK_P(brother->rb_right));
|
||
}
|
||
if (RB_BLACK_P(parent)) {
|
||
/*
|
||
* Both our parent and brother are black.
|
||
* Change our brother to red, advance up rank
|
||
* and go through the loop again.
|
||
*
|
||
* B -> B
|
||
* A D -> A d
|
||
* C E -> C E
|
||
*/
|
||
RB_MARK_RED(brother);
|
||
if (RB_ROOT_P(parent))
|
||
return;
|
||
KASSERT(rb_tree_check_node(rbt, brother, NULL, false));
|
||
KASSERT(rb_tree_check_node(rbt, parent, NULL, false));
|
||
which = parent->rb_position;
|
||
parent = parent->rb_parent;
|
||
} else {
|
||
KASSERT(RB_BLACK_P(brother));
|
||
RB_MARK_BLACK(parent);
|
||
RB_MARK_RED(brother);
|
||
KASSERT(rb_tree_check_node(rbt, brother, NULL, true));
|
||
KASSERT(rb_tree_check_node(rbt, parent, NULL, true));
|
||
return;
|
||
}
|
||
} else {
|
||
KASSERT(RB_BLACK_P(brother));
|
||
KASSERT(!RB_CHILDLESS_P(brother));
|
||
/*
|
||
* Case 3: our brother is black, our left nephew is
|
||
* red, and our right nephew is black. Swap our
|
||
* brother with our left nephew. This result in a
|
||
* tree that matches case 4.
|
||
*
|
||
* B -> D
|
||
* A D -> B E
|
||
* c e -> A C
|
||
*/
|
||
if (RB_BLACK_P(brother->rb_nodes[other])) {
|
||
KASSERT(RB_RED_P(brother->rb_nodes[which]));
|
||
rb_tree_swap_nodes(rbt, brother, which);
|
||
brother = parent->rb_nodes[other];
|
||
KASSERT(RB_RED_P(brother->rb_nodes[other]));
|
||
}
|
||
/*
|
||
* Case 4: our brother is black and our right nephew
|
||
* is red. Swap our parent and brother locations and
|
||
* change our right nephew to black. (these can be
|
||
* done in either order so we change the color first).
|
||
* The result is a valid red-black tree and is a
|
||
* terminal case.
|
||
*
|
||
* B -> D
|
||
* A D -> B E
|
||
* c e -> A C
|
||
*/
|
||
RB_MARK_BLACK(brother->rb_nodes[other]);
|
||
rb_tree_swap_nodes(rbt, parent, other);
|
||
KASSERT(rb_tree_check_node(rbt, parent, NULL, true));
|
||
return;
|
||
}
|
||
}
|
||
KASSERT(rb_tree_check_node(rbt, parent, NULL, true));
|
||
}
|
||
|
||
#ifndef NDEBUG
|
||
static bool
|
||
rb_tree_check_node(const struct rb_tree *rbt, const struct rb_node *self,
|
||
const struct rb_node *prev, bool red_check)
|
||
{
|
||
KASSERT(!self->rb_sentinel);
|
||
KASSERT(self->rb_left);
|
||
KASSERT(self->rb_right);
|
||
KASSERT(prev == NULL || (*rbt->rbt_compare_nodes)(prev, self) > 0);
|
||
|
||
/*
|
||
* Verify our relationship to our parent.
|
||
*/
|
||
if (RB_ROOT_P(self)) {
|
||
KASSERT(self == rbt->rbt_root);
|
||
KASSERT(self->rb_position == RB_LEFT);
|
||
KASSERT(self->rb_parent->rb_nodes[RB_LEFT] == self);
|
||
KASSERT(self->rb_parent == (const struct rb_node *) &rbt->rbt_root);
|
||
} else {
|
||
KASSERT(self != rbt->rbt_root);
|
||
KASSERT(!RB_PARENT_SENTINEL_P(self));
|
||
if (self->rb_position == RB_LEFT) {
|
||
KASSERT((*rbt->rbt_compare_nodes)(self, self->rb_parent) > 0);
|
||
KASSERT(self->rb_parent->rb_nodes[RB_LEFT] == self);
|
||
} else {
|
||
KASSERT((*rbt->rbt_compare_nodes)(self, self->rb_parent) < 0);
|
||
KASSERT(self->rb_parent->rb_nodes[RB_RIGHT] == self);
|
||
}
|
||
}
|
||
|
||
/*
|
||
* The root must be black.
|
||
* There can never be two adjacent red nodes.
|
||
*/
|
||
if (red_check) {
|
||
KASSERT(!RB_ROOT_P(self) || RB_BLACK_P(self));
|
||
if (RB_RED_P(self)) {
|
||
KASSERT(!RB_ROOT_P(self));
|
||
KASSERT(RB_BLACK_P(self->rb_parent));
|
||
KASSERT(RB_LEFT_SENTINEL_P(self)
|
||
|| RB_BLACK_P(self->rb_left));
|
||
KASSERT(RB_RIGHT_SENTINEL_P(self)
|
||
|| RB_BLACK_P(self->rb_right));
|
||
}
|
||
/*
|
||
* A grandparent's children must be real nodes and not
|
||
* sentinels. First check out grandparent.
|
||
*/
|
||
KASSERT(RB_ROOT_P(self)
|
||
|| RB_ROOT_P(self->rb_parent)
|
||
|| RB_TWOCHILDREN_P(self->rb_parent->rb_parent));
|
||
/*
|
||
* If we are have grandchildren on our left, then
|
||
* we must have a child on our right.
|
||
*/
|
||
KASSERT(RB_LEFT_SENTINEL_P(self)
|
||
|| RB_CHILDLESS_P(self->rb_left)
|
||
|| !RB_RIGHT_SENTINEL_P(self));
|
||
/*
|
||
* If we are have grandchildren on our right, then
|
||
* we must have a child on our left.
|
||
*/
|
||
KASSERT(RB_RIGHT_SENTINEL_P(self)
|
||
|| RB_CHILDLESS_P(self->rb_right)
|
||
|| !RB_LEFT_SENTINEL_P(self));
|
||
}
|
||
|
||
return true;
|
||
}
|
||
|
||
void
|
||
rb_tree_check(const struct rb_tree *rbt, bool red_check)
|
||
{
|
||
const struct rb_node *self;
|
||
const struct rb_node *prev;
|
||
int counts[2] = { 0, 0 };
|
||
unsigned int which = RB_LEFT;
|
||
|
||
KASSERT(rbt->rbt_root == NULL || rbt->rbt_root->rb_position == RB_LEFT);
|
||
|
||
prev = NULL;
|
||
TAILQ_FOREACH(self, &rbt->rbt_nodes, rb_link) {
|
||
rb_tree_check_node(rbt, self, prev, false);
|
||
if (self == rbt->rbt_root) {
|
||
which = RB_RIGHT;
|
||
} else {
|
||
counts[which]++;
|
||
}
|
||
prev = self;
|
||
}
|
||
KASSERT(rbt->rbt_count == counts[RB_RIGHT] + counts[RB_LEFT] + (which == RB_RIGHT));
|
||
|
||
/*
|
||
* The root must be black.
|
||
* There can never be two adjacent red nodes.
|
||
*/
|
||
if (red_check) {
|
||
KASSERT(rbt->rbt_root == NULL || RB_BLACK_P(rbt->rbt_root));
|
||
TAILQ_FOREACH(self, &rbt->rbt_nodes, rb_link) {
|
||
rb_tree_check_node(rbt, self, NULL, true);
|
||
}
|
||
}
|
||
}
|
||
#endif
|