2018-03-22 04:31:06 +08:00
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/*
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* Copyright (c) 2018 Intel Corporation
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*
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* SPDX-License-Identifier: Apache-2.0
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*/
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/* These assertions are very useful when debugging the tree code
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* itself, but produce significant performance degradation as they are
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* checked many times per operation. Leave them off unless you're
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* working on the rbtree code itself
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*/
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#define CHECK(n) /**/
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/* #define CHECK(n) __ASSERT_NO_MSG(n) */
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#include <kernel.h>
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#include <misc/rb.h>
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enum rb_color { RED = 0, BLACK = 1 };
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static struct rbnode *get_child(struct rbnode *n, int side)
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{
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CHECK(n);
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if (side) {
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return n->children[1];
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}
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uintptr_t l = (uintptr_t) n->children[0];
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l &= ~1ul;
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return (struct rbnode *) l;
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}
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static void set_child(struct rbnode *n, int side, void *val)
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{
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CHECK(n);
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if (side) {
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n->children[1] = val;
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} else {
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uintptr_t old = (uintptr_t) n->children[0];
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uintptr_t new = (uintptr_t) val;
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n->children[0] = (void *) (new | (old & 1ul));
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}
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}
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static enum rb_color get_color(struct rbnode *n)
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{
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CHECK(n);
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return ((uintptr_t)n->children[0]) & 1ul;
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}
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static int is_black(struct rbnode *n)
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{
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return get_color(n) == BLACK;
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}
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static int is_red(struct rbnode *n)
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{
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return get_color(n) == RED;
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}
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static void set_color(struct rbnode *n, enum rb_color color)
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{
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CHECK(n);
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uintptr_t *p = (void *) &n->children[0];
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*p = (*p & ~1ul) | color;
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}
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/* Searches the tree down to a node that is either identical with the
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* "node" argument or has an empty/leaf child pointer where "node"
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* should be, leaving all nodes found in the resulting stack. Note
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* that tree must not be empty and that stack should be allocated to
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* contain at least tree->max_depth entries! Returns the number of
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* entries pushed onto the stack.
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*/
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static int find_and_stack(struct rbtree *tree, struct rbnode *node,
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struct rbnode **stack)
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{
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int sz = 0;
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stack[sz++] = tree->root;
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while (stack[sz - 1] != node) {
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int side = tree->lessthan_fn(node, stack[sz - 1]) ? 0 : 1;
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struct rbnode *ch = get_child(stack[sz - 1], side);
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if (ch) {
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stack[sz++] = ch;
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} else {
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break;
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}
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}
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return sz;
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}
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struct rbnode *_rb_get_minmax(struct rbtree *tree, int side)
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{
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struct rbnode *n;
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for (n = tree->root; get_child(n, side); n = get_child(n, side))
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;
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return n;
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}
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static int get_side(struct rbnode *parent, struct rbnode *child)
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{
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CHECK(get_child(parent, 0) == child || get_child(parent, 1) == child);
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return get_child(parent, 1) == child ? 1 : 0;
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}
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/* Swaps the position of the two nodes at the top of the provided
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* stack, modifying the stack accordingly. Does not change the color
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* of either node. That is, it effects the following transition (or
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* its mirror if N is on the other side of P, of course):
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*
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* P N
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* N c --> a P
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* a b b c
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*
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*/
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static void rotate(struct rbnode **stack, int stacksz)
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{
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CHECK(stacksz >= 2);
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struct rbnode *parent = stack[stacksz - 2];
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struct rbnode *child = stack[stacksz - 1];
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int side = get_side(parent, child);
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struct rbnode *a = get_child(child, side);
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struct rbnode *b = get_child(child, !side);
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if (stacksz >= 3) {
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struct rbnode *grandparent = stack[stacksz - 3];
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set_child(grandparent, get_side(grandparent, parent), child);
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}
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set_child(child, side, a);
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set_child(child, !side, parent);
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set_child(parent, side, b);
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stack[stacksz - 2] = child;
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stack[stacksz - 1] = parent;
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}
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/* The node at the top of the provided stack is red, and its parent is
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* too. Iteratively fix the tree so it becomes a valid red black tree
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* again
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*/
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static void fix_extra_red(struct rbnode **stack, int stacksz)
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{
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while (stacksz > 1) {
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struct rbnode *node = stack[stacksz - 1];
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struct rbnode *parent = stack[stacksz - 2];
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/* Correct child colors are a precondition of the loop */
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CHECK(!get_child(node, 0) || is_black(get_child(node, 0)));
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CHECK(!get_child(node, 1) || is_black(get_child(node, 1)));
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if (is_black(parent)) {
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return;
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}
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/* We are guaranteed to have a grandparent if our
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* parent is red, as red nodes cannot be the root
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*/
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CHECK(stacksz >= 2);
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struct rbnode *grandparent = stack[stacksz - 3];
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int side = get_side(grandparent, parent);
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struct rbnode *aunt = get_child(grandparent, !side);
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if (aunt && is_red(aunt)) {
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set_color(grandparent, RED);
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set_color(parent, BLACK);
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set_color(aunt, BLACK);
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/* We colored the grandparent red, which might
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* have a red parent, so continue iterating
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* from there.
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*/
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stacksz -= 2;
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continue;
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}
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/* We can rotate locally to fix the whole tree. First
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* make sure that node is on the same side of parent
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* as parent is of grandparent.
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*/
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int parent_side = get_side(parent, node);
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if (parent_side != side) {
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rotate(stack, stacksz);
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node = stack[stacksz - 1];
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}
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/* Rotate the grandparent with parent, swapping colors */
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rotate(stack, stacksz - 1);
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set_color(stack[stacksz - 3], BLACK);
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set_color(stack[stacksz - 2], RED);
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return;
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}
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/* If we exit the loop, it's because our node is now the root,
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* which must be black.
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*/
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set_color(stack[0], BLACK);
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}
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void rb_insert(struct rbtree *tree, struct rbnode *node)
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{
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set_child(node, 0, NULL);
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set_child(node, 1, NULL);
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if (!tree->root) {
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tree->root = node;
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tree->max_depth = 1;
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set_color(node, BLACK);
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return;
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}
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struct rbnode *stack[tree->max_depth + 1];
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int stacksz = find_and_stack(tree, node, stack);
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struct rbnode *parent = stack[stacksz - 1];
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int side = tree->lessthan_fn(parent, node);
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set_child(parent, side, node);
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set_color(node, RED);
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stack[stacksz++] = node;
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fix_extra_red(stack, stacksz);
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if (stacksz > tree->max_depth) {
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tree->max_depth = stacksz;
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}
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/* We may have rotated up into the root! */
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tree->root = stack[0];
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CHECK(is_black(tree->root));
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}
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/* Called for a node N (at the top of the stack) which after a
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* deletion operation is "missing a black" in its subtree. By
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* construction N must be black (because if it was red it would be
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* trivially fixed by recoloring and we wouldn't be here). Fixes up
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* the tree to preserve red/black rules. The "null_node" pointer is
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* for situations where we are removing a childless black node. The
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* tree munging needs a real node for simplicity, so we use it and
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* then clean it up (replace it with a simple NULL child in the
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* parent) when finished.
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*/
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static void fix_missing_black(struct rbnode **stack, int stacksz,
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struct rbnode *null_node)
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{
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/* Loop upward until we reach the root */
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while (stacksz > 1) {
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struct rbnode *c0, *c1, *inner, *outer;
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struct rbnode *n = stack[stacksz - 1];
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struct rbnode *parent = stack[stacksz - 2];
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int n_side = get_side(parent, n);
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struct rbnode *sib = get_child(parent, !n_side);
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CHECK(is_black(n));
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/* Guarantee the sibling is black, rotating N down a
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* level if needed (after rotate() our parent is the
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* child of our previous-sibling, so N is lower in the
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* tree)
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*/
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if (!is_black(sib)) {
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stack[stacksz - 1] = sib;
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rotate(stack, stacksz);
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set_color(parent, RED);
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set_color(sib, BLACK);
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stack[stacksz++] = n;
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parent = stack[stacksz - 2];
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sib = get_child(parent, !n_side);
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}
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CHECK(sib);
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/* Cases where the sibling has only black children
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* have simple resolutions
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*/
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c0 = get_child(sib, 0);
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c1 = get_child(sib, 1);
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if ((!c0 || is_black(c0)) && (!c1 || is_black(c1))) {
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if (n == null_node) {
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set_child(parent, n_side, NULL);
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}
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set_color(sib, RED);
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if (is_black(parent)) {
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/* Balance the sibling's subtree by
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* coloring it red, then our parent
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* has a missing black so iterate
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* upward
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*/
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stacksz--;
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continue;
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} else {
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/* Recoloring makes the whole tree OK */
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set_color(parent, BLACK);
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return;
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}
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}
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CHECK((c0 && is_red(c0)) || (c1 && is_red(c1)));
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/* We know sibling has at least one red child. Fix it
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* so that the far/outer position (i.e. on the
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* opposite side from N) is definitely red.
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*/
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outer = get_child(sib, !n_side);
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if (!(outer && is_red(outer))) {
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inner = get_child(sib, n_side);
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stack[stacksz - 1] = sib;
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stack[stacksz++] = inner;
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rotate(stack, stacksz);
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set_color(sib, RED);
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set_color(inner, BLACK);
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/* Restore stack state to have N on the top
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* and make sib reflect the new sibling
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*/
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sib = stack[stacksz - 2];
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outer = get_child(sib, !n_side);
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stack[stacksz - 2] = n;
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stacksz--;
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}
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/* Finally, the sibling must have a red child in the
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* far/outer slot. We can rotate sib with our parent
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* and recolor to produce a valid tree.
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*/
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CHECK(is_red(outer));
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set_color(sib, get_color(parent));
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set_color(parent, BLACK);
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set_color(outer, BLACK);
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stack[stacksz - 1] = sib;
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rotate(stack, stacksz);
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if (n == null_node) {
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set_child(parent, n_side, NULL);
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}
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return;
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}
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}
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void rb_remove(struct rbtree *tree, struct rbnode *node)
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{
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struct rbnode *stack[tree->max_depth + 1], *tmp;
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int stacksz = find_and_stack(tree, node, stack);
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if (node != stack[stacksz - 1]) {
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return;
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}
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/* We can only remove a node with zero or one child, if we
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* have two then pick the "biggest" child of side 0 (smallest
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* of 1 would work too) and swap our spot in the tree with
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* that one
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*/
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if (get_child(node, 0) && get_child(node, 1)) {
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int stacksz0 = stacksz;
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struct rbnode *hiparent, *loparent;
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struct rbnode *node2 = get_child(node, 0);
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hiparent = stacksz > 1 ? stack[stacksz - 2] : NULL;
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stack[stacksz++] = node2;
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while (get_child(node2, 1)) {
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node2 = get_child(node2, 1);
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stack[stacksz++] = node2;
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}
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loparent = stack[stacksz - 2];
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/* Now swap the position of node/node2 in the tree.
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* Design note: this is a spot where being an
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* intrusive data structure hurts us fairly badly.
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* The trees you see in textbooks do this by swapping
|
|
|
|
* the "data" pointers between the two nodes, but we
|
|
|
|
* have a few special cases to check. In principle
|
|
|
|
* this works by swapping the child pointers between
|
|
|
|
* the nodes and retargetting the nodes pointing to
|
|
|
|
* them from their parents, but: (1) the upper node
|
|
|
|
* may be the root of the tree and not have a parent,
|
|
|
|
* and (2) the lower node may be a direct child of the
|
|
|
|
* upper node. Remember to swap the color bits of the
|
|
|
|
* two nodes also. And of course we don't have parent
|
|
|
|
* pointers, so the stack tracking this structure
|
|
|
|
* needs to be swapped too!
|
|
|
|
*/
|
|
|
|
if (hiparent) {
|
|
|
|
set_child(hiparent, get_side(hiparent, node), node2);
|
|
|
|
} else {
|
|
|
|
tree->root = node2;
|
|
|
|
}
|
|
|
|
|
|
|
|
if (loparent == node) {
|
|
|
|
set_child(node, 0, get_child(node2, 0));
|
|
|
|
set_child(node2, 0, node);
|
|
|
|
} else {
|
|
|
|
set_child(loparent, get_side(loparent, node2), node);
|
|
|
|
tmp = get_child(node, 0);
|
|
|
|
set_child(node, 0, get_child(node2, 0));
|
|
|
|
set_child(node2, 0, tmp);
|
|
|
|
}
|
|
|
|
|
|
|
|
set_child(node2, 1, get_child(node, 1));
|
|
|
|
set_child(node, 1, NULL);
|
|
|
|
|
|
|
|
tmp = stack[stacksz0 - 1];
|
|
|
|
stack[stacksz0 - 1] = stack[stacksz - 1];
|
|
|
|
stack[stacksz - 1] = tmp;
|
|
|
|
|
|
|
|
int ctmp = get_color(node);
|
|
|
|
|
|
|
|
set_color(node, get_color(node2));
|
|
|
|
set_color(node2, ctmp);
|
|
|
|
}
|
|
|
|
|
|
|
|
CHECK(!get_child(node, 0) || !get_child(node, 1));
|
|
|
|
|
|
|
|
struct rbnode *child = get_child(node, 0);
|
|
|
|
|
|
|
|
if (!child) {
|
|
|
|
child = get_child(node, 1);
|
|
|
|
}
|
|
|
|
|
|
|
|
/* Removing the root */
|
|
|
|
if (stacksz < 2) {
|
|
|
|
tree->root = child;
|
|
|
|
if (child) {
|
|
|
|
set_color(child, BLACK);
|
|
|
|
} else {
|
|
|
|
tree->max_depth = 0;
|
|
|
|
}
|
|
|
|
return;
|
|
|
|
}
|
|
|
|
|
|
|
|
struct rbnode *parent = stack[stacksz - 2];
|
|
|
|
|
|
|
|
/* Special case: if the node to be removed is childless, then
|
|
|
|
* we leave it in place while we do the missing black
|
|
|
|
* rotations, which will replace it with a proper NULL when
|
|
|
|
* they isolate it.
|
|
|
|
*/
|
|
|
|
if (!child) {
|
|
|
|
if (is_black(node)) {
|
|
|
|
fix_missing_black(stack, stacksz, node);
|
|
|
|
} else {
|
|
|
|
/* Red childless nodes can just be dropped */
|
|
|
|
set_child(parent, get_side(parent, node), NULL);
|
|
|
|
}
|
|
|
|
} else {
|
|
|
|
set_child(parent, get_side(parent, node), child);
|
|
|
|
|
|
|
|
/* Check colors, if one was red (at least one must have been
|
|
|
|
* black in a valid tree), then we're done. Otherwise we have
|
|
|
|
* a missing black we need to fix
|
|
|
|
*/
|
|
|
|
if (is_red(node) || is_red(child)) {
|
|
|
|
set_color(child, BLACK);
|
|
|
|
} else {
|
|
|
|
stack[stacksz - 1] = child;
|
|
|
|
fix_missing_black(stack, stacksz, NULL);
|
|
|
|
}
|
|
|
|
}
|
|
|
|
|
|
|
|
/* We may have rotated up into the root! */
|
|
|
|
tree->root = stack[0];
|
|
|
|
}
|
|
|
|
|
|
|
|
void _rb_walk(struct rbnode *node, rb_visit_t visit_fn, void *cookie)
|
|
|
|
{
|
|
|
|
if (node) {
|
|
|
|
_rb_walk(get_child(node, 0), visit_fn, cookie);
|
|
|
|
visit_fn(node, cookie);
|
|
|
|
_rb_walk(get_child(node, 1), visit_fn, cookie);
|
|
|
|
}
|
|
|
|
}
|
|
|
|
|
|
|
|
struct rbnode *_rb_child(struct rbnode *node, int side)
|
|
|
|
{
|
|
|
|
return get_child(node, side);
|
|
|
|
}
|
|
|
|
|
|
|
|
int _rb_is_black(struct rbnode *node)
|
|
|
|
{
|
|
|
|
return is_black(node);
|
|
|
|
}
|
2018-04-10 23:54:04 +08:00
|
|
|
|
|
|
|
int rb_contains(struct rbtree *tree, struct rbnode *node)
|
|
|
|
{
|
|
|
|
struct rbnode *n = tree->root;
|
|
|
|
|
|
|
|
while (n && n != node) {
|
|
|
|
n = get_child(n, tree->lessthan_fn(n, node));
|
|
|
|
}
|
|
|
|
|
|
|
|
return n == node;
|
|
|
|
}
|