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) |