1 /* Copyright (C) 1995-1997, 2000, 2006-2007, 2009 Free Software Foundation, Inc.
2 Contributed by Bernd Schmidt <crux@Pool.Informatik.RWTH-Aachen.DE>, 1997.
4 NOTE: The canonical source of this file is maintained with the GNU C
5 Library. Bugs can be reported to bug-glibc@gnu.org.
7 This program is free software: you can redistribute it and/or modify it
8 under the terms of the GNU General Public License as published by the
9 Free Software Foundation; either version 3 of the License, or any
12 This program is distributed in the hope that it will be useful,
13 but WITHOUT ANY WARRANTY; without even the implied warranty of
14 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
15 GNU General Public License for more details.
17 You should have received a copy of the GNU General Public License
18 along with this program. If not, see <http://www.gnu.org/licenses/>. */
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
89 /* Don't use __attribute__ __nonnull__ in this compilation unit. Otherwise gcc
90 optimizes away the rootp == NULL tests below. */
91 #define _GL_ARG_NONNULL(params)
102 typedef int (*__compar_fn_t) (const void *, const void *);
103 typedef void (*__action_fn_t) (const void *, VISIT, int);
106 # define __tsearch tsearch
107 # define __tfind tfind
108 # define __tdelete tdelete
109 # define __twalk twalk
112 #ifndef internal_function
113 /* Inside GNU libc we mark some function in a special way. In other
114 environments simply ignore the marking. */
115 # define internal_function
118 typedef struct node_t
120 /* Callers expect this to be the first element in the structure - do not
124 struct node_t *right;
127 typedef const struct node_t *const_node;
133 /* Routines to check tree invariants. */
137 #define CHECK_TREE(a) check_tree(a)
140 check_tree_recurse (node p, int d_sofar, int d_total)
144 assert (d_sofar == d_total);
148 check_tree_recurse (p->left, d_sofar + (p->left && !p->left->red), d_total);
149 check_tree_recurse (p->right, d_sofar + (p->right && !p->right->red), d_total);
151 assert (!(p->left->red && p->red));
153 assert (!(p->right->red && p->red));
157 check_tree (node root)
164 for(p = root->left; p; p = p->left)
166 check_tree_recurse (root, 0, cnt);
172 #define CHECK_TREE(a)
176 /* Possibly "split" a node with two red successors, and/or fix up two red
177 edges in a row. ROOTP is a pointer to the lowest node we visited, PARENTP
178 and GPARENTP pointers to its parent/grandparent. P_R and GP_R contain the
179 comparison values that determined which way was taken in the tree to reach
180 ROOTP. MODE is 1 if we need not do the split, but must check for two red
181 edges between GPARENTP and ROOTP. */
183 maybe_split_for_insert (node *rootp, node *parentp, node *gparentp,
184 int p_r, int gp_r, int mode)
188 rp = &(*rootp)->right;
189 lp = &(*rootp)->left;
191 /* See if we have to split this node (both successors red). */
193 || ((*rp) != NULL && (*lp) != NULL && (*rp)->red && (*lp)->red))
195 /* This node becomes red, its successors black. */
202 /* If the parent of this node is also red, we have to do
204 if (parentp != NULL && (*parentp)->red)
208 /* There are two main cases:
209 1. The edge types (left or right) of the two red edges differ.
210 2. Both red edges are of the same type.
211 There exist two symmetries of each case, so there is a total of
213 if ((p_r > 0) != (gp_r > 0))
215 /* Put the child at the top of the tree, with its parent
216 and grandparent as successors. */
222 /* Child is left of parent. */
230 /* Child is right of parent. */
240 *gparentp = *parentp;
241 /* Parent becomes the top of the tree, grandparent and
242 child are its successors. */
262 /* Find or insert datum into search tree.
263 KEY is the key to be located, ROOTP is the address of tree root,
264 COMPAR the ordering function. */
266 __tsearch (const void *key, void **vrootp, __compar_fn_t compar)
269 node *parentp = NULL, *gparentp = NULL;
270 node *rootp = (node *) vrootp;
272 int r = 0, p_r = 0, gp_r = 0; /* No they might not, Mr Compiler. */
277 /* This saves some additional tests below. */
284 while (*nextp != NULL)
287 r = (*compar) (key, root->key);
291 maybe_split_for_insert (rootp, parentp, gparentp, p_r, gp_r, 0);
292 /* If that did any rotations, parentp and gparentp are now garbage.
293 That doesn't matter, because the values they contain are never
294 used again in that case. */
296 nextp = r < 0 ? &root->left : &root->right;
308 q = (struct node_t *) malloc (sizeof (struct node_t));
311 *nextp = q; /* link new node to old */
312 q->key = key; /* initialize new node */
314 q->left = q->right = NULL;
317 /* There may be two red edges in a row now, which we must avoid by
318 rotating the tree. */
319 maybe_split_for_insert (nextp, rootp, parentp, r, p_r, 1);
325 weak_alias (__tsearch, tsearch)
329 /* Find datum in search tree.
330 KEY is the key to be located, ROOTP is the address of tree root,
331 COMPAR the ordering function. */
333 __tfind (key, vrootp, compar)
336 __compar_fn_t compar;
338 node *rootp = (node *) vrootp;
345 while (*rootp != NULL)
350 r = (*compar) (key, root->key);
354 rootp = r < 0 ? &root->left : &root->right;
359 weak_alias (__tfind, tfind)
363 /* Delete node with given key.
364 KEY is the key to be deleted, ROOTP is the address of the root of tree,
365 COMPAR the comparison function. */
367 __tdelete (const void *key, void **vrootp, __compar_fn_t compar)
369 node p, q, r, retval;
371 node *rootp = (node *) vrootp;
372 node root, unchained;
373 /* Stack of nodes so we remember the parents without recursion. It's
374 _very_ unlikely that there are paths longer than 40 nodes. The tree
375 would need to have around 250.000 nodes. */
378 node *nodestack[100];
388 while ((cmp = (*compar) (key, (*rootp)->key)) != 0)
393 nodestack[sp++] = rootp;
402 /* This is bogus if the node to be deleted is the root... this routine
403 really should return an integer with 0 for success, -1 for failure
404 and errno = ESRCH or something. */
407 /* We don't unchain the node we want to delete. Instead, we overwrite
408 it with its successor and unchain the successor. If there is no
409 successor, we really unchain the node to be deleted. */
416 if (q == NULL || r == NULL)
420 node *parent = rootp, *up = &root->right;
425 nodestack[sp++] = parent;
427 if ((*up)->left == NULL)
434 /* We know that either the left or right successor of UNCHAINED is NULL.
435 R becomes the other one, it is chained into the parent of UNCHAINED. */
438 r = unchained->right;
443 q = *nodestack[sp-1];
444 if (unchained == q->right)
450 if (unchained != root)
451 root->key = unchained->key;
454 /* Now we lost a black edge, which means that the number of black
455 edges on every path is no longer constant. We must balance the
457 /* NODESTACK now contains all parents of R. R is likely to be NULL
458 in the first iteration. */
459 /* NULL nodes are considered black throughout - this is necessary for
461 while (sp > 0 && (r == NULL || !r->red))
463 node *pp = nodestack[sp - 1];
465 /* Two symmetric cases. */
468 /* Q is R's brother, P is R's parent. The subtree with root
469 R has one black edge less than the subtree with root Q. */
473 /* If Q is red, we know that P is black. We rotate P left
474 so that Q becomes the top node in the tree, with P below
475 it. P is colored red, Q is colored black.
476 This action does not change the black edge count for any
477 leaf in the tree, but we will be able to recognize one
478 of the following situations, which all require that Q
486 /* Make sure pp is right if the case below tries to use
488 nodestack[sp++] = pp = &q->left;
491 /* We know that Q can't be NULL here. We also know that Q is
493 if ((q->left == NULL || !q->left->red)
494 && (q->right == NULL || !q->right->red))
496 /* Q has two black successors. We can simply color Q red.
497 The whole subtree with root P is now missing one black
498 edge. Note that this action can temporarily make the
499 tree invalid (if P is red). But we will exit the loop
500 in that case and set P black, which both makes the tree
501 valid and also makes the black edge count come out
502 right. If P is black, we are at least one step closer
503 to the root and we'll try again the next iteration. */
509 /* Q is black, one of Q's successors is red. We can
510 repair the tree with one operation and will exit the
512 if (q->right == NULL || !q->right->red)
514 /* The left one is red. We perform the same action as
515 in maybe_split_for_insert where two red edges are
516 adjacent but point in different directions:
517 Q's left successor (let's call it Q2) becomes the
518 top of the subtree we are looking at, its parent (Q)
519 and grandparent (P) become its successors. The former
520 successors of Q2 are placed below P and Q.
521 P becomes black, and Q2 gets the color that P had.
522 This changes the black edge count only for node R and
535 /* It's the right one. Rotate P left. P becomes black,
536 and Q gets the color that P had. Q's right successor
537 also becomes black. This changes the black edge
538 count only for node R and its successors. */
557 /* Comments: see above. */
566 nodestack[sp++] = pp = &q->right;
569 if ((q->right == NULL || !q->right->red)
570 && (q->left == NULL || !q->left->red))
577 if (q->left == NULL || !q->left->red)
611 weak_alias (__tdelete, tdelete)
615 /* Walk the nodes of a tree.
616 ROOT is the root of the tree to be walked, ACTION the function to be
617 called at each node. LEVEL is the level of ROOT in the whole tree. */
620 trecurse (const void *vroot, __action_fn_t action, int level)
622 const_node root = (const_node) vroot;
624 if (root->left == NULL && root->right == NULL)
625 (*action) (root, leaf, level);
628 (*action) (root, preorder, level);
629 if (root->left != NULL)
630 trecurse (root->left, action, level + 1);
631 (*action) (root, postorder, level);
632 if (root->right != NULL)
633 trecurse (root->right, action, level + 1);
634 (*action) (root, endorder, level);
639 /* Walk the nodes of a tree.
640 ROOT is the root of the tree to be walked, ACTION the function to be
641 called at each node. */
643 __twalk (const void *vroot, __action_fn_t action)
645 const_node root = (const_node) vroot;
649 if (root != NULL && action != NULL)
650 trecurse (root, action, 0);
653 weak_alias (__twalk, twalk)
659 /* The standardized functions miss an important functionality: the
660 tree cannot be removed easily. We provide a function to do this. */
663 tdestroy_recurse (node root, __free_fn_t freefct)
665 if (root->left != NULL)
666 tdestroy_recurse (root->left, freefct);
667 if (root->right != NULL)
668 tdestroy_recurse (root->right, freefct);
669 (*freefct) ((void *) root->key);
670 /* Free the node itself. */
675 __tdestroy (void *vroot, __free_fn_t freefct)
677 node root = (node) vroot;
682 tdestroy_recurse (root, freefct);
684 weak_alias (__tdestroy, tdestroy)