1 /* Copyright (C) 1995, 1996, 1997, 2000, 2006 Free Software Foundation, Inc.
2 This file is part of the GNU C Library.
3 Contributed by Bernd Schmidt <crux@Pool.Informatik.RWTH-Aachen.DE>, 1997.
5 The GNU C Library is free software; you can redistribute it and/or
6 modify it under the terms of the GNU Lesser General Public
7 License as published by the Free Software Foundation; either
8 version 2.1 of the License, or (at your option) any later version.
10 The GNU C Library is distributed in the hope that it will be useful,
11 but WITHOUT ANY WARRANTY; without even the implied warranty of
12 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
13 Lesser General Public License for more details.
15 You should have received a copy of the GNU Lesser General Public
16 License along with the GNU C Library; if not, write to the Free
17 Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA
20 /* Tree search for red/black trees.
21 The algorithm for adding nodes is taken from one of the many "Algorithms"
22 books by Robert Sedgewick, although the implementation differs.
23 The algorithm for deleting nodes can probably be found in a book named
24 "Introduction to Algorithms" by Cormen/Leiserson/Rivest. At least that's
25 the book that my professor took most algorithms from during the "Data
28 Totally public domain. */
30 /* Red/black trees are binary trees in which the edges are colored either red
31 or black. They have the following properties:
32 1. The number of black edges on every path from the root to a leaf is
34 2. No two red edges are adjacent.
35 Therefore there is an upper bound on the length of every path, it's
36 O(log n) where n is the number of nodes in the tree. No path can be longer
37 than 1+2*P where P is the length of the shortest path in the tree.
38 Useful for the implementation:
39 3. If one of the children of a node is NULL, then the other one is red
42 In the implementation, not the edges are colored, but the nodes. The color
43 interpreted as the color of the edge leading to this node. The color is
44 meaningless for the root node, but we color the root node black for
45 convenience. All added nodes are red initially.
47 Adding to a red/black tree is rather easy. The right place is searched
48 with a usual binary tree search. Additionally, whenever a node N is
49 reached that has two red successors, the successors are colored black and
50 the node itself colored red. This moves red edges up the tree where they
51 pose less of a problem once we get to really insert the new node. Changing
52 N's color to red may violate rule 2, however, so rotations may become
53 necessary to restore the invariants. Adding a new red leaf may violate
54 the same rule, so afterwards an additional check is run and the tree
57 Deleting is hairy. There are mainly two nodes involved: the node to be
58 deleted (n1), and another node that is to be unchained from the tree (n2).
59 If n1 has a successor (the node with a smallest key that is larger than
60 n1), then the successor becomes n2 and its contents are copied into n1,
61 otherwise n1 becomes n2.
62 Unchaining a node may violate rule 1: if n2 is black, one subtree is
63 missing one black edge afterwards. The algorithm must try to move this
64 error upwards towards the root, so that the subtree that does not have
65 enough black edges becomes the whole tree. Once that happens, the error
66 has disappeared. It may not be necessary to go all the way up, since it
67 is possible that rotations and recoloring can fix the error before that.
69 Although the deletion algorithm must walk upwards through the tree, we
70 do not store parent pointers in the nodes. Instead, delete allocates a
71 small array of parent pointers and fills it while descending the tree.
72 Since we know that the length of a path is O(log n), where n is the number
73 of nodes, this is likely to use less memory. */
75 /* Tree rotations look like this:
84 In this case, A has been rotated left. This preserves the ordering of the
93 /* Callers expect this to be the first element in the structure - do not
100 typedef const struct node_t *const_node;
106 /* Routines to check tree invariants. */
110 #define CHECK_TREE(a) check_tree(a)
113 check_tree_recurse (node p, int d_sofar, int d_total)
117 assert (d_sofar == d_total);
121 check_tree_recurse (p->left, d_sofar + (p->left && !p->left->red), d_total);
122 check_tree_recurse (p->right, d_sofar + (p->right && !p->right->red), d_total);
124 assert (!(p->left->red && p->red));
126 assert (!(p->right->red && p->red));
130 check_tree (node root)
137 for(p = root->left; p; p = p->left)
139 check_tree_recurse (root, 0, cnt);
145 #define CHECK_TREE(a)
149 /* Possibly "split" a node with two red successors, and/or fix up two red
150 edges in a row. ROOTP is a pointer to the lowest node we visited, PARENTP
151 and GPARENTP pointers to its parent/grandparent. P_R and GP_R contain the
152 comparison values that determined which way was taken in the tree to reach
153 ROOTP. MODE is 1 if we need not do the split, but must check for two red
154 edges between GPARENTP and ROOTP. */
156 maybe_split_for_insert (node *rootp, node *parentp, node *gparentp,
157 int p_r, int gp_r, int mode)
161 rp = &(*rootp)->right;
162 lp = &(*rootp)->left;
164 /* See if we have to split this node (both successors red). */
166 || ((*rp) != NULL && (*lp) != NULL && (*rp)->red && (*lp)->red))
168 /* This node becomes red, its successors black. */
175 /* If the parent of this node is also red, we have to do
177 if (parentp != NULL && (*parentp)->red)
181 /* There are two main cases:
182 1. The edge types (left or right) of the two red edges differ.
183 2. Both red edges are of the same type.
184 There exist two symmetries of each case, so there is a total of
186 if ((p_r > 0) != (gp_r > 0))
188 /* Put the child at the top of the tree, with its parent
189 and grandparent as successors. */
195 /* Child is left of parent. */
203 /* Child is right of parent. */
213 *gparentp = *parentp;
214 /* Parent becomes the top of the tree, grandparent and
215 child are its successors. */
235 /* Find or insert datum into search tree.
236 KEY is the key to be located, ROOTP is the address of tree root,
237 COMPAR the ordering function. */
239 __tsearch (const void *key, void **vrootp, __compar_fn_t compar)
242 node *parentp = NULL, *gparentp = NULL;
243 node *rootp = (node *) vrootp;
245 int r = 0, p_r = 0, gp_r = 0; /* No they might not, Mr Compiler. */
250 /* This saves some additional tests below. */
257 while (*nextp != NULL)
260 r = (*compar) (key, root->key);
264 maybe_split_for_insert (rootp, parentp, gparentp, p_r, gp_r, 0);
265 /* If that did any rotations, parentp and gparentp are now garbage.
266 That doesn't matter, because the values they contain are never
267 used again in that case. */
269 nextp = r < 0 ? &root->left : &root->right;
281 q = (struct node_t *) malloc (sizeof (struct node_t));
284 *nextp = q; /* link new node to old */
285 q->key = key; /* initialize new node */
287 q->left = q->right = NULL;
290 /* There may be two red edges in a row now, which we must avoid by
291 rotating the tree. */
292 maybe_split_for_insert (nextp, rootp, parentp, r, p_r, 1);
297 weak_alias (__tsearch, tsearch)
300 /* Find datum in search tree.
301 KEY is the key to be located, ROOTP is the address of tree root,
302 COMPAR the ordering function. */
304 __tfind (key, vrootp, compar)
307 __compar_fn_t compar;
309 node *rootp = (node *) vrootp;
316 while (*rootp != NULL)
321 r = (*compar) (key, root->key);
325 rootp = r < 0 ? &root->left : &root->right;
329 weak_alias (__tfind, tfind)
332 /* Delete node with given key.
333 KEY is the key to be deleted, ROOTP is the address of the root of tree,
334 COMPAR the comparison function. */
336 __tdelete (const void *key, void **vrootp, __compar_fn_t compar)
338 node p, q, r, retval;
340 node *rootp = (node *) vrootp;
341 node root, unchained;
342 /* Stack of nodes so we remember the parents without recursion. It's
343 _very_ unlikely that there are paths longer than 40 nodes. The tree
344 would need to have around 250.000 nodes. */
347 node **nodestack = alloca (sizeof (node *) * stacksize);
357 while ((cmp = (*compar) (key, (*rootp)->key)) != 0)
363 newstack = alloca (sizeof (node *) * stacksize);
364 nodestack = memcpy (newstack, nodestack, sp * sizeof (node *));
367 nodestack[sp++] = rootp;
376 /* This is bogus if the node to be deleted is the root... this routine
377 really should return an integer with 0 for success, -1 for failure
378 and errno = ESRCH or something. */
381 /* We don't unchain the node we want to delete. Instead, we overwrite
382 it with its successor and unchain the successor. If there is no
383 successor, we really unchain the node to be deleted. */
390 if (q == NULL || r == NULL)
394 node *parent = rootp, *up = &root->right;
401 newstack = alloca (sizeof (node *) * stacksize);
402 nodestack = memcpy (newstack, nodestack, sp * sizeof (node *));
404 nodestack[sp++] = parent;
406 if ((*up)->left == NULL)
413 /* We know that either the left or right successor of UNCHAINED is NULL.
414 R becomes the other one, it is chained into the parent of UNCHAINED. */
417 r = unchained->right;
422 q = *nodestack[sp-1];
423 if (unchained == q->right)
429 if (unchained != root)
430 root->key = unchained->key;
433 /* Now we lost a black edge, which means that the number of black
434 edges on every path is no longer constant. We must balance the
436 /* NODESTACK now contains all parents of R. R is likely to be NULL
437 in the first iteration. */
438 /* NULL nodes are considered black throughout - this is necessary for
440 while (sp > 0 && (r == NULL || !r->red))
442 node *pp = nodestack[sp - 1];
444 /* Two symmetric cases. */
447 /* Q is R's brother, P is R's parent. The subtree with root
448 R has one black edge less than the subtree with root Q. */
452 /* If Q is red, we know that P is black. We rotate P left
453 so that Q becomes the top node in the tree, with P below
454 it. P is colored red, Q is colored black.
455 This action does not change the black edge count for any
456 leaf in the tree, but we will be able to recognize one
457 of the following situations, which all require that Q
465 /* Make sure pp is right if the case below tries to use
467 nodestack[sp++] = pp = &q->left;
470 /* We know that Q can't be NULL here. We also know that Q is
472 if ((q->left == NULL || !q->left->red)
473 && (q->right == NULL || !q->right->red))
475 /* Q has two black successors. We can simply color Q red.
476 The whole subtree with root P is now missing one black
477 edge. Note that this action can temporarily make the
478 tree invalid (if P is red). But we will exit the loop
479 in that case and set P black, which both makes the tree
480 valid and also makes the black edge count come out
481 right. If P is black, we are at least one step closer
482 to the root and we'll try again the next iteration. */
488 /* Q is black, one of Q's successors is red. We can
489 repair the tree with one operation and will exit the
491 if (q->right == NULL || !q->right->red)
493 /* The left one is red. We perform the same action as
494 in maybe_split_for_insert where two red edges are
495 adjacent but point in different directions:
496 Q's left successor (let's call it Q2) becomes the
497 top of the subtree we are looking at, its parent (Q)
498 and grandparent (P) become its successors. The former
499 successors of Q2 are placed below P and Q.
500 P becomes black, and Q2 gets the color that P had.
501 This changes the black edge count only for node R and
514 /* It's the right one. Rotate P left. P becomes black,
515 and Q gets the color that P had. Q's right successor
516 also becomes black. This changes the black edge
517 count only for node R and its successors. */
536 /* Comments: see above. */
545 nodestack[sp++] = pp = &q->right;
548 if ((q->right == NULL || !q->right->red)
549 && (q->left == NULL || !q->left->red))
556 if (q->left == NULL || !q->left->red)
589 weak_alias (__tdelete, tdelete)
592 /* Walk the nodes of a tree.
593 ROOT is the root of the tree to be walked, ACTION the function to be
594 called at each node. LEVEL is the level of ROOT in the whole tree. */
597 trecurse (const void *vroot, __action_fn_t action, int level)
599 const_node root = (const_node) vroot;
601 if (root->left == NULL && root->right == NULL)
602 (*action) (root, leaf, level);
605 (*action) (root, preorder, level);
606 if (root->left != NULL)
607 trecurse (root->left, action, level + 1);
608 (*action) (root, postorder, level);
609 if (root->right != NULL)
610 trecurse (root->right, action, level + 1);
611 (*action) (root, endorder, level);
616 /* Walk the nodes of a tree.
617 ROOT is the root of the tree to be walked, ACTION the function to be
618 called at each node. */
620 __twalk (const void *vroot, __action_fn_t action)
622 const_node root = (const_node) vroot;
626 if (root != NULL && action != NULL)
627 trecurse (root, action, 0);
629 weak_alias (__twalk, twalk)
633 /* The standardized functions miss an important functionality: the
634 tree cannot be removed easily. We provide a function to do this. */
637 tdestroy_recurse (node root, __free_fn_t freefct)
639 if (root->left != NULL)
640 tdestroy_recurse (root->left, freefct);
641 if (root->right != NULL)
642 tdestroy_recurse (root->right, freefct);
643 (*freefct) ((void *) root->key);
644 /* Free the node itself. */
649 __tdestroy (void *vroot, __free_fn_t freefct)
651 node root = (node) vroot;
656 tdestroy_recurse (root, freefct);
658 weak_alias (__tdestroy, tdestroy)