| xf.li | bdd93d5 | 2023-05-12 07:10:14 -0700 | [diff] [blame] | 1 | /* Copyright (C) 1995-2016 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. | 
|  | 4 |  | 
|  | 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. | 
|  | 9 |  | 
|  | 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. | 
|  | 14 |  | 
|  | 15 | You should have received a copy of the GNU Lesser General Public | 
|  | 16 | License along with the GNU C Library; if not, see | 
|  | 17 | <http://www.gnu.org/licenses/>.  */ | 
|  | 18 |  | 
|  | 19 | /* Tree search for red/black trees. | 
|  | 20 | The algorithm for adding nodes is taken from one of the many "Algorithms" | 
|  | 21 | books by Robert Sedgewick, although the implementation differs. | 
|  | 22 | The algorithm for deleting nodes can probably be found in a book named | 
|  | 23 | "Introduction to Algorithms" by Cormen/Leiserson/Rivest.  At least that's | 
|  | 24 | the book that my professor took most algorithms from during the "Data | 
|  | 25 | Structures" course... | 
|  | 26 |  | 
|  | 27 | Totally public domain.  */ | 
|  | 28 |  | 
|  | 29 | /* Red/black trees are binary trees in which the edges are colored either red | 
|  | 30 | or black.  They have the following properties: | 
|  | 31 | 1. The number of black edges on every path from the root to a leaf is | 
|  | 32 | constant. | 
|  | 33 | 2. No two red edges are adjacent. | 
|  | 34 | Therefore there is an upper bound on the length of every path, it's | 
|  | 35 | O(log n) where n is the number of nodes in the tree.  No path can be longer | 
|  | 36 | than 1+2*P where P is the length of the shortest path in the tree. | 
|  | 37 | Useful for the implementation: | 
|  | 38 | 3. If one of the children of a node is NULL, then the other one is red | 
|  | 39 | (if it exists). | 
|  | 40 |  | 
|  | 41 | In the implementation, not the edges are colored, but the nodes.  The color | 
|  | 42 | interpreted as the color of the edge leading to this node.  The color is | 
|  | 43 | meaningless for the root node, but we color the root node black for | 
|  | 44 | convenience.  All added nodes are red initially. | 
|  | 45 |  | 
|  | 46 | Adding to a red/black tree is rather easy.  The right place is searched | 
|  | 47 | with a usual binary tree search.  Additionally, whenever a node N is | 
|  | 48 | reached that has two red successors, the successors are colored black and | 
|  | 49 | the node itself colored red.  This moves red edges up the tree where they | 
|  | 50 | pose less of a problem once we get to really insert the new node.  Changing | 
|  | 51 | N's color to red may violate rule 2, however, so rotations may become | 
|  | 52 | necessary to restore the invariants.  Adding a new red leaf may violate | 
|  | 53 | the same rule, so afterwards an additional check is run and the tree | 
|  | 54 | possibly rotated. | 
|  | 55 |  | 
|  | 56 | Deleting is hairy.  There are mainly two nodes involved: the node to be | 
|  | 57 | deleted (n1), and another node that is to be unchained from the tree (n2). | 
|  | 58 | If n1 has a successor (the node with a smallest key that is larger than | 
|  | 59 | n1), then the successor becomes n2 and its contents are copied into n1, | 
|  | 60 | otherwise n1 becomes n2. | 
|  | 61 | Unchaining a node may violate rule 1: if n2 is black, one subtree is | 
|  | 62 | missing one black edge afterwards.  The algorithm must try to move this | 
|  | 63 | error upwards towards the root, so that the subtree that does not have | 
|  | 64 | enough black edges becomes the whole tree.  Once that happens, the error | 
|  | 65 | has disappeared.  It may not be necessary to go all the way up, since it | 
|  | 66 | is possible that rotations and recoloring can fix the error before that. | 
|  | 67 |  | 
|  | 68 | Although the deletion algorithm must walk upwards through the tree, we | 
|  | 69 | do not store parent pointers in the nodes.  Instead, delete allocates a | 
|  | 70 | small array of parent pointers and fills it while descending the tree. | 
|  | 71 | Since we know that the length of a path is O(log n), where n is the number | 
|  | 72 | of nodes, this is likely to use less memory.  */ | 
|  | 73 |  | 
|  | 74 | /* Tree rotations look like this: | 
|  | 75 | A                C | 
|  | 76 | / \              / \ | 
|  | 77 | B   C            A   G | 
|  | 78 | / \ / \  -->     / \ | 
|  | 79 | D E F G         B   F | 
|  | 80 | / \ | 
|  | 81 | D   E | 
|  | 82 |  | 
|  | 83 | In this case, A has been rotated left.  This preserves the ordering of the | 
|  | 84 | binary tree.  */ | 
|  | 85 |  | 
|  | 86 | #include <stdlib.h> | 
|  | 87 | #include <string.h> | 
|  | 88 | #include <search.h> | 
|  | 89 |  | 
|  | 90 | typedef struct node_t | 
|  | 91 | { | 
|  | 92 | /* Callers expect this to be the first element in the structure - do not | 
|  | 93 | move!  */ | 
|  | 94 | const void *key; | 
|  | 95 | struct node_t *left; | 
|  | 96 | struct node_t *right; | 
|  | 97 | unsigned int red:1; | 
|  | 98 | } *node; | 
|  | 99 | typedef const struct node_t *const_node; | 
|  | 100 |  | 
|  | 101 | #undef DEBUGGING | 
|  | 102 |  | 
|  | 103 | #ifdef DEBUGGING | 
|  | 104 |  | 
|  | 105 | /* Routines to check tree invariants.  */ | 
|  | 106 |  | 
|  | 107 | #include <assert.h> | 
|  | 108 |  | 
|  | 109 | #define CHECK_TREE(a) check_tree(a) | 
|  | 110 |  | 
|  | 111 | static void | 
|  | 112 | check_tree_recurse (node p, int d_sofar, int d_total) | 
|  | 113 | { | 
|  | 114 | if (p == NULL) | 
|  | 115 | { | 
|  | 116 | assert (d_sofar == d_total); | 
|  | 117 | return; | 
|  | 118 | } | 
|  | 119 |  | 
|  | 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)); | 
|  | 126 | } | 
|  | 127 |  | 
|  | 128 | static void | 
|  | 129 | check_tree (node root) | 
|  | 130 | { | 
|  | 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); | 
|  | 139 | } | 
|  | 140 |  | 
|  | 141 |  | 
|  | 142 | #else | 
|  | 143 |  | 
|  | 144 | #define CHECK_TREE(a) | 
|  | 145 |  | 
|  | 146 | #endif | 
|  | 147 |  | 
|  | 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) | 
|  | 157 | { | 
|  | 158 | node root = *rootp; | 
|  | 159 | node *rp, *lp; | 
|  | 160 | rp = &(*rootp)->right; | 
|  | 161 | lp = &(*rootp)->left; | 
|  | 162 |  | 
|  | 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)) | 
|  | 166 | { | 
|  | 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; | 
|  | 173 |  | 
|  | 174 | /* If the parent of this node is also red, we have to do | 
|  | 175 | rotations.  */ | 
|  | 176 | if (parentp != NULL && (*parentp)->red) | 
|  | 177 | { | 
|  | 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)) | 
|  | 186 | { | 
|  | 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) | 
|  | 193 | { | 
|  | 194 | /* Child is left of parent.  */ | 
|  | 195 | p->left = *rp; | 
|  | 196 | *rp = p; | 
|  | 197 | gp->right = *lp; | 
|  | 198 | *lp = gp; | 
|  | 199 | } | 
|  | 200 | else | 
|  | 201 | { | 
|  | 202 | /* Child is right of parent.  */ | 
|  | 203 | p->right = *lp; | 
|  | 204 | *lp = p; | 
|  | 205 | gp->left = *rp; | 
|  | 206 | *rp = gp; | 
|  | 207 | } | 
|  | 208 | *gparentp = root; | 
|  | 209 | } | 
|  | 210 | else | 
|  | 211 | { | 
|  | 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) | 
|  | 218 | { | 
|  | 219 | /* Left edges.  */ | 
|  | 220 | gp->left = p->right; | 
|  | 221 | p->right = gp; | 
|  | 222 | } | 
|  | 223 | else | 
|  | 224 | { | 
|  | 225 | /* Right edges.  */ | 
|  | 226 | gp->right = p->left; | 
|  | 227 | p->left = gp; | 
|  | 228 | } | 
|  | 229 | } | 
|  | 230 | } | 
|  | 231 | } | 
|  | 232 | } | 
|  | 233 |  | 
|  | 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) | 
|  | 239 | { | 
|  | 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.  */ | 
|  | 245 |  | 
|  | 246 | if (rootp == NULL) | 
|  | 247 | return NULL; | 
|  | 248 |  | 
|  | 249 | /* This saves some additional tests below.  */ | 
|  | 250 | if (*rootp != NULL) | 
|  | 251 | (*rootp)->red = 0; | 
|  | 252 |  | 
|  | 253 | CHECK_TREE (*rootp); | 
|  | 254 |  | 
|  | 255 | nextp = rootp; | 
|  | 256 | while (*nextp != NULL) | 
|  | 257 | { | 
|  | 258 | node root = *rootp; | 
|  | 259 | r = (*compar) (key, root->key); | 
|  | 260 | if (r == 0) | 
|  | 261 | return root; | 
|  | 262 |  | 
|  | 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.  */ | 
|  | 267 |  | 
|  | 268 | nextp = r < 0 ? &root->left : &root->right; | 
|  | 269 | if (*nextp == NULL) | 
|  | 270 | break; | 
|  | 271 |  | 
|  | 272 | gparentp = parentp; | 
|  | 273 | parentp = rootp; | 
|  | 274 | rootp = nextp; | 
|  | 275 |  | 
|  | 276 | gp_r = p_r; | 
|  | 277 | p_r = r; | 
|  | 278 | } | 
|  | 279 |  | 
|  | 280 | q = (struct node_t *) malloc (sizeof (struct node_t)); | 
|  | 281 | if (q != NULL) | 
|  | 282 | { | 
|  | 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; | 
|  | 287 |  | 
|  | 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); | 
|  | 292 | } | 
|  | 293 |  | 
|  | 294 | return q; | 
|  | 295 | } | 
|  | 296 | libc_hidden_def (__tsearch) | 
|  | 297 | weak_alias (__tsearch, tsearch) | 
|  | 298 |  | 
|  | 299 |  | 
|  | 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.  */ | 
|  | 303 | void * | 
|  | 304 | __tfind (const void *key, void *const *vrootp, __compar_fn_t compar) | 
|  | 305 | { | 
|  | 306 | node *rootp = (node *) vrootp; | 
|  | 307 |  | 
|  | 308 | if (rootp == NULL) | 
|  | 309 | return NULL; | 
|  | 310 |  | 
|  | 311 | CHECK_TREE (*rootp); | 
|  | 312 |  | 
|  | 313 | while (*rootp != NULL) | 
|  | 314 | { | 
|  | 315 | node root = *rootp; | 
|  | 316 | int r; | 
|  | 317 |  | 
|  | 318 | r = (*compar) (key, root->key); | 
|  | 319 | if (r == 0) | 
|  | 320 | return root; | 
|  | 321 |  | 
|  | 322 | rootp = r < 0 ? &root->left : &root->right; | 
|  | 323 | } | 
|  | 324 | return NULL; | 
|  | 325 | } | 
|  | 326 | libc_hidden_def (__tfind) | 
|  | 327 | weak_alias (__tfind, tfind) | 
|  | 328 |  | 
|  | 329 |  | 
|  | 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) | 
|  | 335 | { | 
|  | 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); | 
|  | 346 |  | 
|  | 347 | if (rootp == NULL) | 
|  | 348 | return NULL; | 
|  | 349 | p = *rootp; | 
|  | 350 | if (p == NULL) | 
|  | 351 | return NULL; | 
|  | 352 |  | 
|  | 353 | CHECK_TREE (p); | 
|  | 354 |  | 
|  | 355 | while ((cmp = (*compar) (key, (*rootp)->key)) != 0) | 
|  | 356 | { | 
|  | 357 | if (sp == stacksize) | 
|  | 358 | { | 
|  | 359 | node **newstack; | 
|  | 360 | stacksize += 20; | 
|  | 361 | newstack = alloca (sizeof (node *) * stacksize); | 
|  | 362 | nodestack = memcpy (newstack, nodestack, sp * sizeof (node *)); | 
|  | 363 | } | 
|  | 364 |  | 
|  | 365 | nodestack[sp++] = rootp; | 
|  | 366 | p = *rootp; | 
|  | 367 | rootp = ((cmp < 0) | 
|  | 368 | ? &(*rootp)->left | 
|  | 369 | : &(*rootp)->right); | 
|  | 370 | if (*rootp == NULL) | 
|  | 371 | return NULL; | 
|  | 372 | } | 
|  | 373 |  | 
|  | 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; | 
|  | 378 |  | 
|  | 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.  */ | 
|  | 382 |  | 
|  | 383 | root = *rootp; | 
|  | 384 |  | 
|  | 385 | r = root->right; | 
|  | 386 | q = root->left; | 
|  | 387 |  | 
|  | 388 | if (q == NULL || r == NULL) | 
|  | 389 | unchained = root; | 
|  | 390 | else | 
|  | 391 | { | 
|  | 392 | node *parent = rootp, *up = &root->right; | 
|  | 393 | for (;;) | 
|  | 394 | { | 
|  | 395 | if (sp == stacksize) | 
|  | 396 | { | 
|  | 397 | node **newstack; | 
|  | 398 | stacksize += 20; | 
|  | 399 | newstack = alloca (sizeof (node *) * stacksize); | 
|  | 400 | nodestack = memcpy (newstack, nodestack, sp * sizeof (node *)); | 
|  | 401 | } | 
|  | 402 | nodestack[sp++] = parent; | 
|  | 403 | parent = up; | 
|  | 404 | if ((*up)->left == NULL) | 
|  | 405 | break; | 
|  | 406 | up = &(*up)->left; | 
|  | 407 | } | 
|  | 408 | unchained = *up; | 
|  | 409 | } | 
|  | 410 |  | 
|  | 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 | 
|  | 419 | { | 
|  | 420 | q = *nodestack[sp-1]; | 
|  | 421 | if (unchained == q->right) | 
|  | 422 | q->right = r; | 
|  | 423 | else | 
|  | 424 | q->left = r; | 
|  | 425 | } | 
|  | 426 |  | 
|  | 427 | if (unchained != root) | 
|  | 428 | root->key = unchained->key; | 
|  | 429 | if (!unchained->red) | 
|  | 430 | { | 
|  | 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)) | 
|  | 439 | { | 
|  | 440 | node *pp = nodestack[sp - 1]; | 
|  | 441 | p = *pp; | 
|  | 442 | /* Two symmetric cases.  */ | 
|  | 443 | if (r == p->left) | 
|  | 444 | { | 
|  | 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->red) | 
|  | 449 | { | 
|  | 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; | 
|  | 467 | } | 
|  | 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)) | 
|  | 472 | { | 
|  | 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; | 
|  | 483 | } | 
|  | 484 | else | 
|  | 485 | { | 
|  | 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) | 
|  | 490 | { | 
|  | 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; | 
|  | 509 | } | 
|  | 510 | else | 
|  | 511 | { | 
|  | 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; | 
|  | 518 |  | 
|  | 519 | q->right->red = 0; | 
|  | 520 |  | 
|  | 521 | /* left rotate p */ | 
|  | 522 | p->right = q->left; | 
|  | 523 | q->left = p; | 
|  | 524 | *pp = q; | 
|  | 525 | } | 
|  | 526 |  | 
|  | 527 | /* We're done.  */ | 
|  | 528 | sp = 1; | 
|  | 529 | r = NULL; | 
|  | 530 | } | 
|  | 531 | } | 
|  | 532 | else | 
|  | 533 | { | 
|  | 534 | /* Comments: see above.  */ | 
|  | 535 | q = p->left; | 
|  | 536 | if (q->red) | 
|  | 537 | { | 
|  | 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; | 
|  | 545 | } | 
|  | 546 | if ((q->right == NULL || !q->right->red) | 
|  | 547 | && (q->left == NULL || !q->left->red)) | 
|  | 548 | { | 
|  | 549 | q->red = 1; | 
|  | 550 | r = p; | 
|  | 551 | } | 
|  | 552 | else | 
|  | 553 | { | 
|  | 554 | if (q->left == NULL || !q->left->red) | 
|  | 555 | { | 
|  | 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; | 
|  | 564 | } | 
|  | 565 | else | 
|  | 566 | { | 
|  | 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; | 
|  | 573 | } | 
|  | 574 | sp = 1; | 
|  | 575 | r = NULL; | 
|  | 576 | } | 
|  | 577 | } | 
|  | 578 | --sp; | 
|  | 579 | } | 
|  | 580 | if (r != NULL) | 
|  | 581 | r->red = 0; | 
|  | 582 | } | 
|  | 583 |  | 
|  | 584 | free (unchained); | 
|  | 585 | return retval; | 
|  | 586 | } | 
|  | 587 | libc_hidden_def (__tdelete) | 
|  | 588 | weak_alias (__tdelete, tdelete) | 
|  | 589 |  | 
|  | 590 |  | 
|  | 591 | /* Walk the nodes of a tree. | 
|  | 592 | ROOT is the root of the tree to be walked, ACTION the function to be | 
|  | 593 | called at each node.  LEVEL is the level of ROOT in the whole tree.  */ | 
|  | 594 | static void | 
|  | 595 | internal_function | 
|  | 596 | trecurse (const void *vroot, __action_fn_t action, int level) | 
|  | 597 | { | 
|  | 598 | const_node root = (const_node) vroot; | 
|  | 599 |  | 
|  | 600 | if (root->left == NULL && root->right == NULL) | 
|  | 601 | (*action) (root, leaf, level); | 
|  | 602 | else | 
|  | 603 | { | 
|  | 604 | (*action) (root, preorder, level); | 
|  | 605 | if (root->left != NULL) | 
|  | 606 | trecurse (root->left, action, level + 1); | 
|  | 607 | (*action) (root, postorder, level); | 
|  | 608 | if (root->right != NULL) | 
|  | 609 | trecurse (root->right, action, level + 1); | 
|  | 610 | (*action) (root, endorder, level); | 
|  | 611 | } | 
|  | 612 | } | 
|  | 613 |  | 
|  | 614 |  | 
|  | 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.  */ | 
|  | 618 | void | 
|  | 619 | __twalk (const void *vroot, __action_fn_t action) | 
|  | 620 | { | 
|  | 621 | const_node root = (const_node) vroot; | 
|  | 622 |  | 
|  | 623 | CHECK_TREE (root); | 
|  | 624 |  | 
|  | 625 | if (root != NULL && action != NULL) | 
|  | 626 | trecurse (root, action, 0); | 
|  | 627 | } | 
|  | 628 | libc_hidden_def (__twalk) | 
|  | 629 | weak_alias (__twalk, twalk) | 
|  | 630 |  | 
|  | 631 |  | 
|  | 632 |  | 
|  | 633 | /* The standardized functions miss an important functionality: the | 
|  | 634 | tree cannot be removed easily.  We provide a function to do this.  */ | 
|  | 635 | static void | 
|  | 636 | internal_function | 
|  | 637 | tdestroy_recurse (node root, __free_fn_t freefct) | 
|  | 638 | { | 
|  | 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.  */ | 
|  | 645 | free (root); | 
|  | 646 | } | 
|  | 647 |  | 
|  | 648 | void | 
|  | 649 | __tdestroy (void *vroot, __free_fn_t freefct) | 
|  | 650 | { | 
|  | 651 | node root = (node) vroot; | 
|  | 652 |  | 
|  | 653 | CHECK_TREE (root); | 
|  | 654 |  | 
|  | 655 | if (root != NULL) | 
|  | 656 | tdestroy_recurse (root, freefct); | 
|  | 657 | } | 
|  | 658 | weak_alias (__tdestroy, tdestroy) |