Update.
[glibc.git] / misc / tsearch.c
blob0cf854b8bc5bbc6eb08497ab89accc3f4fceafcf
1 /* Copyright (C) 1995, 1996, 1997 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 Library General Public License as
7 published by the Free Software Foundation; either version 2 of the
8 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 Library General Public License for more details.
15 You should have received a copy of the GNU Library General Public
16 License along with the GNU C Library; see the file COPYING.LIB. If not,
17 write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330,
18 Boston, MA 02111-1307, USA. */
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
26 Structures" course...
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
33 constant.
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
40 (if it exists).
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
55 possibly rotated.
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:
76 A C
77 / \ / \
78 B C A G
79 / \ / \ --> / \
80 D E F G B F
81 / \
82 D E
84 In this case, A has been rotated left. This preserves the ordering of the
85 binary tree. */
87 #include <stdlib.h>
88 #include <string.h>
89 #include <search.h>
91 typedef struct node_t
93 /* Callers expect this to be the first element in the structure - do not
94 move! */
95 const void *key;
96 struct node_t *left;
97 struct node_t *right;
98 unsigned int red:1;
99 } *node;
101 #undef DEBUGGING
103 #ifdef DEBUGGING
105 /* Routines to check tree invariants. */
107 #include <assert.h>
109 #define CHECK_TREE(a) check_tree(a)
111 static void
112 check_tree_recurse (node p, int d_sofar, int d_total)
114 if (p == NULL)
116 assert (d_sofar == d_total);
117 return;
120 check_tree_recurse (p->left, d_sofar + (p->left && !p->left->red), d_total);
121 check_tree_recurse (p->right, d_sofar + (p->right && !p->right->red), d_total);
122 if (p->left)
123 assert (!(p->left->red && p->red));
124 if (p->right)
125 assert (!(p->right->red && p->red));
128 static void
129 check_tree (node root)
131 int cnt = 0;
132 node p;
133 if (root == NULL)
134 return;
135 root->red = 0;
136 for(p = root->left; p; p = p->left)
137 cnt += !p->red;
138 check_tree_recurse (root, 0, cnt);
142 #else
144 #define CHECK_TREE(a)
146 #endif
148 /* Possibly "split" a node with two red successors, and/or fix up two red
149 edges in a row. ROOTP is a pointer to the lowest node we visited, PARENTP
150 and GPARENTP pointers to its parent/grandparent. P_R and GP_R contain the
151 comparison values that determined which way was taken in the tree to reach
152 ROOTP. MODE is 1 if we need not do the split, but must check for two red
153 edges between GPARENTP and ROOTP. */
154 static void
155 maybe_split_for_insert (node *rootp, node *parentp, node *gparentp,
156 int p_r, int gp_r, int mode)
158 node root = *rootp;
159 node *rp, *lp;
160 rp = &(*rootp)->right;
161 lp = &(*rootp)->left;
163 /* See if we have to split this node (both successors red). */
164 if (mode == 1
165 || ((*rp) != NULL && (*lp) != NULL && (*rp)->red && (*lp)->red))
167 /* This node becomes red, its successors black. */
168 root->red = 1;
169 if (*rp)
170 (*rp)->red = 0;
171 if (*lp)
172 (*lp)->red = 0;
174 /* If the parent of this node is also red, we have to do
175 rotations. */
176 if (parentp != NULL && (*parentp)->red)
178 node gp = *gparentp;
179 node p = *parentp;
180 /* There are two main cases:
181 1. The edge types (left or right) of the two red edges differ.
182 2. Both red edges are of the same type.
183 There exist two symmetries of each case, so there is a total of
184 4 cases. */
185 if ((p_r > 0) != (gp_r > 0))
187 /* Put the child at the top of the tree, with its parent
188 and grandparent as successors. */
189 p->red = 1;
190 gp->red = 1;
191 root->red = 0;
192 if (p_r < 0)
194 /* Child is left of parent. */
195 p->left = *rp;
196 *rp = p;
197 gp->right = *lp;
198 *lp = gp;
200 else
202 /* Child is right of parent. */
203 p->right = *lp;
204 *lp = p;
205 gp->left = *rp;
206 *rp = gp;
208 *gparentp = root;
210 else
212 *gparentp = *parentp;
213 /* Parent becomes the top of the tree, grandparent and
214 child are its successors. */
215 p->red = 0;
216 gp->red = 1;
217 if (p_r < 0)
219 /* Left edges. */
220 gp->left = p->right;
221 p->right = gp;
223 else
225 /* Right edges. */
226 gp->right = p->left;
227 p->left = gp;
234 /* Find or insert datum into search tree.
235 KEY is the key to be located, ROOTP is the address of tree root,
236 COMPAR the ordering function. */
237 void *
238 __tsearch (const void *key, void **vrootp, __compar_fn_t compar)
240 node q;
241 node *parentp = NULL, *gparentp = NULL;
242 node *rootp = (node *) vrootp;
243 node *nextp;
244 int r = 0, p_r = 0, gp_r = 0; /* No they might not, Mr Compiler. */
246 if (rootp == NULL)
247 return NULL;
249 /* This saves some additional tests below. */
250 if (*rootp != NULL)
251 (*rootp)->red = 0;
253 CHECK_TREE (*rootp);
255 nextp = rootp;
256 while (*nextp != NULL)
258 node root = *rootp;
259 r = (*compar) (key, root->key);
260 if (r == 0)
261 return root;
263 maybe_split_for_insert (rootp, parentp, gparentp, p_r, gp_r, 0);
264 /* If that did any rotations, parentp and gparentp are now garbage.
265 That doesn't matter, because the values they contain are never
266 used again in that case. */
268 nextp = r < 0 ? &root->left : &root->right;
269 if (*nextp == NULL)
270 break;
272 gparentp = parentp;
273 parentp = rootp;
274 rootp = nextp;
276 gp_r = p_r;
277 p_r = r;
280 q = (struct node_t *) malloc (sizeof (struct node_t));
281 if (q != NULL)
283 *nextp = q; /* link new node to old */
284 q->key = key; /* initialize new node */
285 q->red = 1;
286 q->left = q->right = NULL;
288 if (nextp != rootp)
289 /* There may be two red edges in a row now, which we must avoid by
290 rotating the tree. */
291 maybe_split_for_insert (nextp, rootp, parentp, r, p_r, 1);
293 return q;
295 weak_alias (__tsearch, tsearch)
298 /* Find datum in search tree.
299 KEY is the key to be located, ROOTP is the address of tree root,
300 COMPAR the ordering function. */
301 void *
302 __tfind (key, vrootp, compar)
303 const void *key;
304 void *const *vrootp;
305 __compar_fn_t compar;
307 node *rootp = (node *) vrootp;
309 if (rootp == NULL)
310 return NULL;
312 CHECK_TREE (*rootp);
314 while (*rootp != NULL)
316 node root = *rootp;
317 int r;
319 r = (*compar) (key, root->key);
320 if (r == 0)
321 return root;
323 rootp = r < 0 ? &root->left : &root->right;
325 return NULL;
327 weak_alias (__tfind, tfind)
330 /* Delete node with given key.
331 KEY is the key to be deleted, ROOTP is the address of the root of tree,
332 COMPAR the comparison function. */
333 void *
334 __tdelete (const void *key, void **vrootp, __compar_fn_t compar)
336 node p, q, r, retval;
337 int cmp;
338 node *rootp = (node *) vrootp;
339 node root, unchained;
340 /* Stack of nodes so we remember the parents without recursion. It's
341 _very_ unlikely that there are paths longer than 40 nodes. The tree
342 would need to have around 250.000 nodes. */
343 int stacksize = 40;
344 int sp = 0;
345 node **nodestack = alloca (sizeof (node *) * stacksize);
347 if (rootp == NULL)
348 return NULL;
349 p = *rootp;
350 if (p == NULL)
351 return NULL;
353 CHECK_TREE (p);
355 while ((cmp = (*compar) (key, (*rootp)->key)) != 0)
357 if (sp == stacksize)
359 node **newstack;
360 stacksize += 20;
361 newstack = alloca (sizeof (node *) * stacksize);
362 nodestack = memcpy (newstack, nodestack, sp * sizeof (node *));
365 nodestack[sp++] = rootp;
366 p = *rootp;
367 rootp = ((cmp < 0)
368 ? &(*rootp)->left
369 : &(*rootp)->right);
370 if (*rootp == NULL)
371 return NULL;
374 /* This is bogus if the node to be deleted is the root... this routine
375 really should return an integer with 0 for success, -1 for failure
376 and errno = ESRCH or something. */
377 retval = p;
379 /* We don't unchain the node we want to delete. Instead, we overwrite
380 it with its successor and unchain the successor. If there is no
381 successor, we really unchain the node to be deleted. */
383 root = *rootp;
385 r = root->right;
386 q = root->left;
388 if (q == NULL || r == NULL)
389 unchained = root;
390 else
392 node *parent = rootp, *up = &root->right;
393 for (;;)
395 if (sp == stacksize)
397 node **newstack;
398 stacksize += 20;
399 newstack = alloca (sizeof (node *) * stacksize);
400 nodestack = memcpy (newstack, nodestack, sp * sizeof (node *));
402 nodestack[sp++] = parent;
403 parent = up;
404 if ((*up)->left == NULL)
405 break;
406 up = &(*up)->left;
408 unchained = *up;
411 /* We know that either the left or right successor of UNCHAINED is NULL.
412 R becomes the other one, it is chained into the parent of UNCHAINED. */
413 r = unchained->left;
414 if (r == NULL)
415 r = unchained->right;
416 if (sp == 0)
417 *rootp = r;
418 else
420 q = *nodestack[sp-1];
421 if (unchained == q->right)
422 q->right = r;
423 else
424 q->left = r;
427 if (unchained != root)
428 root->key = unchained->key;
429 if (!unchained->red)
431 /* Now we lost a black edge, which means that the number of black
432 edges on every path is no longer constant. We must balance the
433 tree. */
434 /* NODESTACK now contains all parents of R. R is likely to be NULL
435 in the first iteration. */
436 /* NULL nodes are considered black throughout - this is necessary for
437 correctness. */
438 while (sp > 0 && (r == NULL || !r->red))
440 node *pp = nodestack[sp - 1];
441 p = *pp;
442 /* Two symmetric cases. */
443 if (r == p->left)
445 /* Q is R's brother, P is R's parent. The subtree with root
446 R has one black edge less than the subtree with root Q. */
447 q = p->right;
448 if (q != NULL && q->red)
450 /* If Q is red, we know that P is black. We rotate P left
451 so that Q becomes the top node in the tree, with P below
452 it. P is colored red, Q is colored black.
453 This action does not change the black edge count for any
454 leaf in the tree, but we will be able to recognize one
455 of the following situations, which all require that Q
456 is black. */
457 q->red = 0;
458 p->red = 1;
459 /* Left rotate p. */
460 p->right = q->left;
461 q->left = p;
462 *pp = q;
463 /* Make sure pp is right if the case below tries to use
464 it. */
465 nodestack[sp++] = pp = &q->left;
466 q = p->right;
468 /* We know that Q can't be NULL here. We also know that Q is
469 black. */
470 if ((q->left == NULL || !q->left->red)
471 && (q->right == NULL || !q->right->red))
473 /* Q has two black successors. We can simply color Q red.
474 The whole subtree with root P is now missing one black
475 edge. Note that this action can temporarily make the
476 tree invalid (if P is red). But we will exit the loop
477 in that case and set P black, which both makes the tree
478 valid and also makes the black edge count come out
479 right. If P is black, we are at least one step closer
480 to the root and we'll try again the next iteration. */
481 q->red = 1;
482 r = p;
484 else
486 /* Q is black, one of Q's successors is red. We can
487 repair the tree with one operation and will exit the
488 loop afterwards. */
489 if (q->right == NULL || !q->right->red)
491 /* The left one is red. We perform the same action as
492 in maybe_split_for_insert where two red edges are
493 adjacent but point in different directions:
494 Q's left successor (let's call it Q2) becomes the
495 top of the subtree we are looking at, its parent (Q)
496 and grandparent (P) become its successors. The former
497 successors of Q2 are placed below P and Q.
498 P becomes black, and Q2 gets the color that P had.
499 This changes the black edge count only for node R and
500 its successors. */
501 node q2 = q->left;
502 q2->red = p->red;
503 p->right = q2->left;
504 q->left = q2->right;
505 q2->right = q;
506 q2->left = p;
507 *pp = q2;
508 p->red = 0;
510 else
512 /* It's the right one. Rotate P left. P becomes black,
513 and Q gets the color that P had. Q's right successor
514 also becomes black. This changes the black edge
515 count only for node R and its successors. */
516 q->red = p->red;
517 p->red = 0;
519 q->right->red = 0;
521 /* left rotate p */
522 p->right = q->left;
523 q->left = p;
524 *pp = q;
527 /* We're done. */
528 sp = 1;
529 r = NULL;
532 else
534 /* Comments: see above. */
535 q = p->left;
536 if (q != NULL && q->red)
538 q->red = 0;
539 p->red = 1;
540 p->left = q->right;
541 q->right = p;
542 *pp = q;
543 nodestack[sp++] = pp = &q->right;
544 q = p->left;
546 if ((q->right == NULL || !q->right->red)
547 && (q->left == NULL || !q->left->red))
549 q->red = 1;
550 r = p;
552 else
554 if (q->left == NULL || !q->left->red)
556 node q2 = q->right;
557 q2->red = p->red;
558 p->left = q2->right;
559 q->right = q2->left;
560 q2->left = q;
561 q2->right = p;
562 *pp = q2;
563 p->red = 0;
565 else
567 q->red = p->red;
568 p->red = 0;
569 q->left->red = 0;
570 p->left = q->right;
571 q->right = p;
572 *pp = q;
574 sp = 1;
575 r = NULL;
578 --sp;
580 if (r != NULL)
581 r->red = 0;
584 free (unchained);
585 return retval;
587 weak_alias (__tdelete, tdelete)
590 /* Walk the nodes of a tree.
591 ROOT is the root of the tree to be walked, ACTION the function to be
592 called at each node. LEVEL is the level of ROOT in the whole tree. */
593 static void
594 internal_function
595 trecurse (const void *vroot, __action_fn_t action, int level)
597 node root = (node ) vroot;
599 if (root->left == NULL && root->right == NULL)
600 (*action) (root, leaf, level);
601 else
603 (*action) (root, preorder, level);
604 if (root->left != NULL)
605 trecurse (root->left, action, level + 1);
606 (*action) (root, postorder, level);
607 if (root->right != NULL)
608 trecurse (root->right, action, level + 1);
609 (*action) (root, endorder, level);
614 /* Walk the nodes of a tree.
615 ROOT is the root of the tree to be walked, ACTION the function to be
616 called at each node. */
617 void
618 __twalk (const void *vroot, __action_fn_t action)
620 const node root = (node) vroot;
622 CHECK_TREE (root);
624 if (root != NULL && action != NULL)
625 trecurse (root, action, 0);
627 weak_alias (__twalk, twalk)
631 /* The standardized functions miss an important functionality: the
632 tree cannot be removed easily. We provide a function to do this. */
633 static void
634 internal_function
635 tdestroy_recurse (node root, __free_fn_t freefct)
637 if (root->left != NULL)
638 tdestroy_recurse (root->left, freefct);
639 if (root->right != NULL)
640 tdestroy_recurse (root->right, freefct);
641 (*freefct) ((void *) root->key);
642 /* Free the node itself. */
643 free (root);
646 void
647 __tdestroy (void *vroot, __free_fn_t freefct)
649 node root = (node) vroot;
651 CHECK_TREE (root);
653 if (root != NULL)
654 tdestroy_recurse (root, freefct);
656 weak_alias (__tdestroy, tdestroy)