| 1 | /* |
| 2 | * tree234.c: reasonably generic 2-3-4 tree routines. Currently |
| 3 | * supports insert, delete, find and iterate operations. |
| 4 | */ |
| 5 | |
| 6 | #include <stdio.h> |
| 7 | #include <stdlib.h> |
| 8 | |
| 9 | #include "tree234.h" |
| 10 | |
| 11 | #define mknew(typ) ( (typ *) malloc (sizeof (typ)) ) |
| 12 | #define sfree free |
| 13 | |
| 14 | #ifdef TEST |
| 15 | #define LOG(x) (printf x) |
| 16 | #else |
| 17 | #define LOG(x) |
| 18 | #endif |
| 19 | |
| 20 | struct tree234_Tag { |
| 21 | node234 *root; |
| 22 | cmpfn234 cmp; |
| 23 | }; |
| 24 | |
| 25 | struct node234_Tag { |
| 26 | node234 *parent; |
| 27 | node234 *kids[4]; |
| 28 | void *elems[3]; |
| 29 | }; |
| 30 | |
| 31 | /* |
| 32 | * Create a 2-3-4 tree. |
| 33 | */ |
| 34 | tree234 *newtree234(cmpfn234 cmp) { |
| 35 | tree234 *ret = mknew(tree234); |
| 36 | LOG(("created tree %p\n", ret)); |
| 37 | ret->root = NULL; |
| 38 | ret->cmp = cmp; |
| 39 | return ret; |
| 40 | } |
| 41 | |
| 42 | /* |
| 43 | * Free a 2-3-4 tree (not including freeing the elements). |
| 44 | */ |
| 45 | static void freenode234(node234 *n) { |
| 46 | if (!n) |
| 47 | return; |
| 48 | freenode234(n->kids[0]); |
| 49 | freenode234(n->kids[1]); |
| 50 | freenode234(n->kids[2]); |
| 51 | freenode234(n->kids[3]); |
| 52 | sfree(n); |
| 53 | } |
| 54 | void freetree234(tree234 *t) { |
| 55 | freenode234(t->root); |
| 56 | sfree(t); |
| 57 | } |
| 58 | |
| 59 | /* |
| 60 | * Add an element e to a 2-3-4 tree t. Returns e on success, or if |
| 61 | * an existing element compares equal, returns that. |
| 62 | */ |
| 63 | void *add234(tree234 *t, void *e) { |
| 64 | node234 *n, **np, *left, *right; |
| 65 | void *orig_e = e; |
| 66 | int c; |
| 67 | |
| 68 | LOG(("adding node %p to tree %p\n", e, t)); |
| 69 | if (t->root == NULL) { |
| 70 | t->root = mknew(node234); |
| 71 | t->root->elems[1] = t->root->elems[2] = NULL; |
| 72 | t->root->kids[0] = t->root->kids[1] = NULL; |
| 73 | t->root->kids[2] = t->root->kids[3] = NULL; |
| 74 | t->root->parent = NULL; |
| 75 | t->root->elems[0] = e; |
| 76 | LOG((" created root %p\n", t->root)); |
| 77 | return orig_e; |
| 78 | } |
| 79 | |
| 80 | np = &t->root; |
| 81 | while (*np) { |
| 82 | n = *np; |
| 83 | LOG((" node %p: %p [%p] %p [%p] %p [%p] %p\n", |
| 84 | n, n->kids[0], n->elems[0], n->kids[1], n->elems[1], |
| 85 | n->kids[2], n->elems[2], n->kids[3])); |
| 86 | if ((c = t->cmp(e, n->elems[0])) < 0) |
| 87 | np = &n->kids[0]; |
| 88 | else if (c == 0) |
| 89 | return n->elems[0]; /* already exists */ |
| 90 | else if (n->elems[1] == NULL || (c = t->cmp(e, n->elems[1])) < 0) |
| 91 | np = &n->kids[1]; |
| 92 | else if (c == 0) |
| 93 | return n->elems[1]; /* already exists */ |
| 94 | else if (n->elems[2] == NULL || (c = t->cmp(e, n->elems[2])) < 0) |
| 95 | np = &n->kids[2]; |
| 96 | else if (c == 0) |
| 97 | return n->elems[2]; /* already exists */ |
| 98 | else |
| 99 | np = &n->kids[3]; |
| 100 | LOG((" moving to child %d (%p)\n", np - n->kids, *np)); |
| 101 | } |
| 102 | |
| 103 | /* |
| 104 | * We need to insert the new element in n at position np. |
| 105 | */ |
| 106 | left = NULL; |
| 107 | right = NULL; |
| 108 | while (n) { |
| 109 | LOG((" at %p: %p [%p] %p [%p] %p [%p] %p\n", |
| 110 | n, n->kids[0], n->elems[0], n->kids[1], n->elems[1], |
| 111 | n->kids[2], n->elems[2], n->kids[3])); |
| 112 | LOG((" need to insert %p [%p] %p at position %d\n", |
| 113 | left, e, right, np - n->kids)); |
| 114 | if (n->elems[1] == NULL) { |
| 115 | /* |
| 116 | * Insert in a 2-node; simple. |
| 117 | */ |
| 118 | if (np == &n->kids[0]) { |
| 119 | LOG((" inserting on left of 2-node\n")); |
| 120 | n->kids[2] = n->kids[1]; |
| 121 | n->elems[1] = n->elems[0]; |
| 122 | n->kids[1] = right; |
| 123 | n->elems[0] = e; |
| 124 | n->kids[0] = left; |
| 125 | } else { /* np == &n->kids[1] */ |
| 126 | LOG((" inserting on right of 2-node\n")); |
| 127 | n->kids[2] = right; |
| 128 | n->elems[1] = e; |
| 129 | n->kids[1] = left; |
| 130 | } |
| 131 | if (n->kids[0]) n->kids[0]->parent = n; |
| 132 | if (n->kids[1]) n->kids[1]->parent = n; |
| 133 | if (n->kids[2]) n->kids[2]->parent = n; |
| 134 | LOG((" done\n")); |
| 135 | break; |
| 136 | } else if (n->elems[2] == NULL) { |
| 137 | /* |
| 138 | * Insert in a 3-node; simple. |
| 139 | */ |
| 140 | if (np == &n->kids[0]) { |
| 141 | LOG((" inserting on left of 3-node\n")); |
| 142 | n->kids[3] = n->kids[2]; |
| 143 | n->elems[2] = n->elems[1]; |
| 144 | n->kids[2] = n->kids[1]; |
| 145 | n->elems[1] = n->elems[0]; |
| 146 | n->kids[1] = right; |
| 147 | n->elems[0] = e; |
| 148 | n->kids[0] = left; |
| 149 | } else if (np == &n->kids[1]) { |
| 150 | LOG((" inserting in middle of 3-node\n")); |
| 151 | n->kids[3] = n->kids[2]; |
| 152 | n->elems[2] = n->elems[1]; |
| 153 | n->kids[2] = right; |
| 154 | n->elems[1] = e; |
| 155 | n->kids[1] = left; |
| 156 | } else { /* np == &n->kids[2] */ |
| 157 | LOG((" inserting on right of 3-node\n")); |
| 158 | n->kids[3] = right; |
| 159 | n->elems[2] = e; |
| 160 | n->kids[2] = left; |
| 161 | } |
| 162 | if (n->kids[0]) n->kids[0]->parent = n; |
| 163 | if (n->kids[1]) n->kids[1]->parent = n; |
| 164 | if (n->kids[2]) n->kids[2]->parent = n; |
| 165 | if (n->kids[3]) n->kids[3]->parent = n; |
| 166 | LOG((" done\n")); |
| 167 | break; |
| 168 | } else { |
| 169 | node234 *m = mknew(node234); |
| 170 | m->parent = n->parent; |
| 171 | LOG((" splitting a 4-node; created new node %p\n", m)); |
| 172 | /* |
| 173 | * Insert in a 4-node; split into a 2-node and a |
| 174 | * 3-node, and move focus up a level. |
| 175 | * |
| 176 | * I don't think it matters which way round we put the |
| 177 | * 2 and the 3. For simplicity, we'll put the 3 first |
| 178 | * always. |
| 179 | */ |
| 180 | if (np == &n->kids[0]) { |
| 181 | m->kids[0] = left; |
| 182 | m->elems[0] = e; |
| 183 | m->kids[1] = right; |
| 184 | m->elems[1] = n->elems[0]; |
| 185 | m->kids[2] = n->kids[1]; |
| 186 | e = n->elems[1]; |
| 187 | n->kids[0] = n->kids[2]; |
| 188 | n->elems[0] = n->elems[2]; |
| 189 | n->kids[1] = n->kids[3]; |
| 190 | } else if (np == &n->kids[1]) { |
| 191 | m->kids[0] = n->kids[0]; |
| 192 | m->elems[0] = n->elems[0]; |
| 193 | m->kids[1] = left; |
| 194 | m->elems[1] = e; |
| 195 | m->kids[2] = right; |
| 196 | e = n->elems[1]; |
| 197 | n->kids[0] = n->kids[2]; |
| 198 | n->elems[0] = n->elems[2]; |
| 199 | n->kids[1] = n->kids[3]; |
| 200 | } else if (np == &n->kids[2]) { |
| 201 | m->kids[0] = n->kids[0]; |
| 202 | m->elems[0] = n->elems[0]; |
| 203 | m->kids[1] = n->kids[1]; |
| 204 | m->elems[1] = n->elems[1]; |
| 205 | m->kids[2] = left; |
| 206 | /* e = e; */ |
| 207 | n->kids[0] = right; |
| 208 | n->elems[0] = n->elems[2]; |
| 209 | n->kids[1] = n->kids[3]; |
| 210 | } else { /* np == &n->kids[3] */ |
| 211 | m->kids[0] = n->kids[0]; |
| 212 | m->elems[0] = n->elems[0]; |
| 213 | m->kids[1] = n->kids[1]; |
| 214 | m->elems[1] = n->elems[1]; |
| 215 | m->kids[2] = n->kids[2]; |
| 216 | n->kids[0] = left; |
| 217 | n->elems[0] = e; |
| 218 | n->kids[1] = right; |
| 219 | e = n->elems[2]; |
| 220 | } |
| 221 | m->kids[3] = n->kids[3] = n->kids[2] = NULL; |
| 222 | m->elems[2] = n->elems[2] = n->elems[1] = NULL; |
| 223 | if (m->kids[0]) m->kids[0]->parent = m; |
| 224 | if (m->kids[1]) m->kids[1]->parent = m; |
| 225 | if (m->kids[2]) m->kids[2]->parent = m; |
| 226 | if (n->kids[0]) n->kids[0]->parent = n; |
| 227 | if (n->kids[1]) n->kids[1]->parent = n; |
| 228 | LOG((" left (%p): %p [%p] %p [%p] %p\n", m, |
| 229 | m->kids[0], m->elems[0], |
| 230 | m->kids[1], m->elems[1], |
| 231 | m->kids[2])); |
| 232 | LOG((" right (%p): %p [%p] %p\n", n, |
| 233 | n->kids[0], n->elems[0], |
| 234 | n->kids[1])); |
| 235 | left = m; |
| 236 | right = n; |
| 237 | } |
| 238 | if (n->parent) |
| 239 | np = (n->parent->kids[0] == n ? &n->parent->kids[0] : |
| 240 | n->parent->kids[1] == n ? &n->parent->kids[1] : |
| 241 | n->parent->kids[2] == n ? &n->parent->kids[2] : |
| 242 | &n->parent->kids[3]); |
| 243 | n = n->parent; |
| 244 | } |
| 245 | |
| 246 | /* |
| 247 | * If we've come out of here by `break', n will still be |
| 248 | * non-NULL and we've finished. If we've come here because n is |
| 249 | * NULL, we need to create a new root for the tree because the |
| 250 | * old one has just split into two. |
| 251 | */ |
| 252 | if (!n) { |
| 253 | LOG((" root is overloaded, split into two\n")); |
| 254 | t->root = mknew(node234); |
| 255 | t->root->kids[0] = left; |
| 256 | t->root->elems[0] = e; |
| 257 | t->root->kids[1] = right; |
| 258 | t->root->elems[1] = NULL; |
| 259 | t->root->kids[2] = NULL; |
| 260 | t->root->elems[2] = NULL; |
| 261 | t->root->kids[3] = NULL; |
| 262 | t->root->parent = NULL; |
| 263 | if (t->root->kids[0]) t->root->kids[0]->parent = t->root; |
| 264 | if (t->root->kids[1]) t->root->kids[1]->parent = t->root; |
| 265 | LOG((" new root is %p [%p] %p\n", |
| 266 | t->root->kids[0], t->root->elems[0], t->root->kids[1])); |
| 267 | } |
| 268 | |
| 269 | return orig_e; |
| 270 | } |
| 271 | |
| 272 | /* |
| 273 | * Find an element e in a 2-3-4 tree t. Returns NULL if not found. |
| 274 | * e is always passed as the first argument to cmp, so cmp can be |
| 275 | * an asymmetric function if desired. cmp can also be passed as |
| 276 | * NULL, in which case the compare function from the tree proper |
| 277 | * will be used. |
| 278 | */ |
| 279 | void *find234(tree234 *t, void *e, cmpfn234 cmp) { |
| 280 | node234 *n; |
| 281 | int c; |
| 282 | |
| 283 | if (t->root == NULL) |
| 284 | return NULL; |
| 285 | |
| 286 | if (cmp == NULL) |
| 287 | cmp = t->cmp; |
| 288 | |
| 289 | n = t->root; |
| 290 | while (n) { |
| 291 | if ( (c = cmp(e, n->elems[0])) < 0) |
| 292 | n = n->kids[0]; |
| 293 | else if (c == 0) |
| 294 | return n->elems[0]; |
| 295 | else if (n->elems[1] == NULL || (c = cmp(e, n->elems[1])) < 0) |
| 296 | n = n->kids[1]; |
| 297 | else if (c == 0) |
| 298 | return n->elems[1]; |
| 299 | else if (n->elems[2] == NULL || (c = cmp(e, n->elems[2])) < 0) |
| 300 | n = n->kids[2]; |
| 301 | else if (c == 0) |
| 302 | return n->elems[2]; |
| 303 | else |
| 304 | n = n->kids[3]; |
| 305 | } |
| 306 | |
| 307 | /* |
| 308 | * We've found our way to the bottom of the tree and we know |
| 309 | * where we would insert this node if we wanted to. But it |
| 310 | * isn't there. |
| 311 | */ |
| 312 | return NULL; |
| 313 | } |
| 314 | |
| 315 | /* |
| 316 | * Delete an element e in a 2-3-4 tree. Does not free the element, |
| 317 | * merely removes all links to it from the tree nodes. |
| 318 | */ |
| 319 | void del234(tree234 *t, void *e) { |
| 320 | node234 *n; |
| 321 | int ei = -1; |
| 322 | |
| 323 | n = t->root; |
| 324 | LOG(("deleting %p from tree %p\n", e, t)); |
| 325 | while (1) { |
| 326 | while (n) { |
| 327 | int c; |
| 328 | int ki; |
| 329 | node234 *sub; |
| 330 | |
| 331 | LOG((" node %p: %p [%p] %p [%p] %p [%p] %p\n", |
| 332 | n, n->kids[0], n->elems[0], n->kids[1], n->elems[1], |
| 333 | n->kids[2], n->elems[2], n->kids[3])); |
| 334 | if ((c = t->cmp(e, n->elems[0])) < 0) { |
| 335 | ki = 0; |
| 336 | } else if (c == 0) { |
| 337 | ei = 0; break; |
| 338 | } else if (n->elems[1] == NULL || (c = t->cmp(e, n->elems[1])) < 0) { |
| 339 | ki = 1; |
| 340 | } else if (c == 0) { |
| 341 | ei = 1; break; |
| 342 | } else if (n->elems[2] == NULL || (c = t->cmp(e, n->elems[2])) < 0) { |
| 343 | ki = 2; |
| 344 | } else if (c == 0) { |
| 345 | ei = 2; break; |
| 346 | } else { |
| 347 | ki = 3; |
| 348 | } |
| 349 | /* |
| 350 | * Recurse down to subtree ki. If it has only one element, |
| 351 | * we have to do some transformation to start with. |
| 352 | */ |
| 353 | LOG((" moving to subtree %d\n", ki)); |
| 354 | sub = n->kids[ki]; |
| 355 | if (!sub->elems[1]) { |
| 356 | LOG((" subtree has only one element!\n", ki)); |
| 357 | if (ki > 0 && n->kids[ki-1]->elems[1]) { |
| 358 | /* |
| 359 | * Case 3a, left-handed variant. Child ki has |
| 360 | * only one element, but child ki-1 has two or |
| 361 | * more. So we need to move a subtree from ki-1 |
| 362 | * to ki. |
| 363 | * |
| 364 | * . C . . B . |
| 365 | * / \ -> / \ |
| 366 | * [more] a A b B c d D e [more] a A b c C d D e |
| 367 | */ |
| 368 | node234 *sib = n->kids[ki-1]; |
| 369 | int lastelem = (sib->elems[2] ? 2 : |
| 370 | sib->elems[1] ? 1 : 0); |
| 371 | sub->kids[2] = sub->kids[1]; |
| 372 | sub->elems[1] = sub->elems[0]; |
| 373 | sub->kids[1] = sub->kids[0]; |
| 374 | sub->elems[0] = n->elems[ki-1]; |
| 375 | sub->kids[0] = sib->kids[lastelem+1]; |
| 376 | if (sub->kids[0]) sub->kids[0]->parent = sub; |
| 377 | n->elems[ki-1] = sib->elems[lastelem]; |
| 378 | sib->kids[lastelem+1] = NULL; |
| 379 | sib->elems[lastelem] = NULL; |
| 380 | LOG((" case 3a left\n")); |
| 381 | } else if (ki < 3 && n->kids[ki+1] && |
| 382 | n->kids[ki+1]->elems[1]) { |
| 383 | /* |
| 384 | * Case 3a, right-handed variant. ki has only |
| 385 | * one element but ki+1 has two or more. Move a |
| 386 | * subtree from ki+1 to ki. |
| 387 | * |
| 388 | * . B . . C . |
| 389 | * / \ -> / \ |
| 390 | * a A b c C d D e [more] a A b B c d D e [more] |
| 391 | */ |
| 392 | node234 *sib = n->kids[ki+1]; |
| 393 | int j; |
| 394 | sub->elems[1] = n->elems[ki]; |
| 395 | sub->kids[2] = sib->kids[0]; |
| 396 | if (sub->kids[2]) sub->kids[2]->parent = sub; |
| 397 | n->elems[ki] = sib->elems[0]; |
| 398 | sib->kids[0] = sib->kids[1]; |
| 399 | for (j = 0; j < 2 && sib->elems[j+1]; j++) { |
| 400 | sib->kids[j+1] = sib->kids[j+2]; |
| 401 | sib->elems[j] = sib->elems[j+1]; |
| 402 | } |
| 403 | sib->kids[j+1] = NULL; |
| 404 | sib->elems[j] = NULL; |
| 405 | LOG((" case 3a right\n")); |
| 406 | } else { |
| 407 | /* |
| 408 | * Case 3b. ki has only one element, and has no |
| 409 | * neighbour with more than one. So pick a |
| 410 | * neighbour and merge it with ki, taking an |
| 411 | * element down from n to go in the middle. |
| 412 | * |
| 413 | * . B . . |
| 414 | * / \ -> | |
| 415 | * a A b c C d a A b B c C d |
| 416 | * |
| 417 | * (Since at all points we have avoided |
| 418 | * descending to a node with only one element, |
| 419 | * we can be sure that n is not reduced to |
| 420 | * nothingness by this move, _unless_ it was |
| 421 | * the very first node, ie the root of the |
| 422 | * tree. In that case we remove the now-empty |
| 423 | * root and replace it with its single large |
| 424 | * child as shown.) |
| 425 | */ |
| 426 | node234 *sib; |
| 427 | int j; |
| 428 | |
| 429 | if (ki > 0) |
| 430 | ki--; |
| 431 | sib = n->kids[ki]; |
| 432 | sub = n->kids[ki+1]; |
| 433 | |
| 434 | sub->kids[3] = sub->kids[1]; |
| 435 | sub->elems[2] = sub->elems[0]; |
| 436 | sub->kids[2] = sub->kids[0]; |
| 437 | sub->elems[1] = n->elems[ki]; |
| 438 | sub->kids[1] = sib->kids[1]; |
| 439 | if (sub->kids[1]) sub->kids[1]->parent = sub; |
| 440 | sub->elems[0] = sib->elems[0]; |
| 441 | sub->kids[0] = sib->kids[0]; |
| 442 | if (sub->kids[0]) sub->kids[0]->parent = sub; |
| 443 | |
| 444 | sfree(sib); |
| 445 | |
| 446 | /* |
| 447 | * That's built the big node in sub. Now we |
| 448 | * need to remove the reference to sib in n. |
| 449 | */ |
| 450 | for (j = ki; j < 3 && n->kids[j+1]; j++) { |
| 451 | n->kids[j] = n->kids[j+1]; |
| 452 | n->elems[j] = j<2 ? n->elems[j+1] : NULL; |
| 453 | } |
| 454 | n->kids[j] = NULL; |
| 455 | if (j < 3) n->elems[j] = NULL; |
| 456 | LOG((" case 3b\n")); |
| 457 | |
| 458 | if (!n->elems[0]) { |
| 459 | /* |
| 460 | * The root is empty and needs to be |
| 461 | * removed. |
| 462 | */ |
| 463 | LOG((" shifting root!\n")); |
| 464 | t->root = sub; |
| 465 | sub->parent = NULL; |
| 466 | sfree(n); |
| 467 | } |
| 468 | } |
| 469 | } |
| 470 | n = sub; |
| 471 | } |
| 472 | if (ei==-1) |
| 473 | return; /* nothing to do; `already removed' */ |
| 474 | |
| 475 | /* |
| 476 | * Treat special case: this is the one remaining item in |
| 477 | * the tree. n is the tree root (no parent), has one |
| 478 | * element (no elems[1]), and has no kids (no kids[0]). |
| 479 | */ |
| 480 | if (!n->parent && !n->elems[1] && !n->kids[0]) { |
| 481 | LOG((" removed last element in tree\n")); |
| 482 | sfree(n); |
| 483 | t->root = NULL; |
| 484 | return; |
| 485 | } |
| 486 | |
| 487 | /* |
| 488 | * Now we have the element we want, as n->elems[ei], and we |
| 489 | * have also arranged for that element not to be the only |
| 490 | * one in its node. So... |
| 491 | */ |
| 492 | |
| 493 | if (!n->kids[0] && n->elems[1]) { |
| 494 | /* |
| 495 | * Case 1. n is a leaf node with more than one element, |
| 496 | * so it's _really easy_. Just delete the thing and |
| 497 | * we're done. |
| 498 | */ |
| 499 | int i; |
| 500 | LOG((" case 1\n")); |
| 501 | for (i = ei; i < 2 && n->elems[i+1]; i++) |
| 502 | n->elems[i] = n->elems[i+1]; |
| 503 | n->elems[i] = NULL; |
| 504 | return; /* finished! */ |
| 505 | } else if (n->kids[ei]->elems[1]) { |
| 506 | /* |
| 507 | * Case 2a. n is an internal node, and the root of the |
| 508 | * subtree to the left of e has more than one element. |
| 509 | * So find the predecessor p to e (ie the largest node |
| 510 | * in that subtree), place it where e currently is, and |
| 511 | * then start the deletion process over again on the |
| 512 | * subtree with p as target. |
| 513 | */ |
| 514 | node234 *m = n->kids[ei]; |
| 515 | void *target; |
| 516 | LOG((" case 2a\n")); |
| 517 | while (m->kids[0]) { |
| 518 | m = (m->kids[3] ? m->kids[3] : |
| 519 | m->kids[2] ? m->kids[2] : |
| 520 | m->kids[1] ? m->kids[1] : m->kids[0]); |
| 521 | } |
| 522 | target = (m->elems[2] ? m->elems[2] : |
| 523 | m->elems[1] ? m->elems[1] : m->elems[0]); |
| 524 | n->elems[ei] = target; |
| 525 | n = n->kids[ei]; |
| 526 | e = target; |
| 527 | } else if (n->kids[ei+1]->elems[1]) { |
| 528 | /* |
| 529 | * Case 2b, symmetric to 2a but s/left/right/ and |
| 530 | * s/predecessor/successor/. (And s/largest/smallest/). |
| 531 | */ |
| 532 | node234 *m = n->kids[ei+1]; |
| 533 | void *target; |
| 534 | LOG((" case 2b\n")); |
| 535 | while (m->kids[0]) { |
| 536 | m = m->kids[0]; |
| 537 | } |
| 538 | target = m->elems[0]; |
| 539 | n->elems[ei] = target; |
| 540 | n = n->kids[ei+1]; |
| 541 | e = target; |
| 542 | } else { |
| 543 | /* |
| 544 | * Case 2c. n is an internal node, and the subtrees to |
| 545 | * the left and right of e both have only one element. |
| 546 | * So combine the two subnodes into a single big node |
| 547 | * with their own elements on the left and right and e |
| 548 | * in the middle, then restart the deletion process on |
| 549 | * that subtree, with e still as target. |
| 550 | */ |
| 551 | node234 *a = n->kids[ei], *b = n->kids[ei+1]; |
| 552 | int j; |
| 553 | |
| 554 | LOG((" case 2c\n")); |
| 555 | a->elems[1] = n->elems[ei]; |
| 556 | a->kids[2] = b->kids[0]; |
| 557 | if (a->kids[2]) a->kids[2]->parent = a; |
| 558 | a->elems[2] = b->elems[0]; |
| 559 | a->kids[3] = b->kids[1]; |
| 560 | if (a->kids[3]) a->kids[3]->parent = a; |
| 561 | sfree(b); |
| 562 | /* |
| 563 | * That's built the big node in a, and destroyed b. Now |
| 564 | * remove the reference to b (and e) in n. |
| 565 | */ |
| 566 | for (j = ei; j < 2 && n->elems[j+1]; j++) { |
| 567 | n->elems[j] = n->elems[j+1]; |
| 568 | n->kids[j+1] = n->kids[j+2]; |
| 569 | } |
| 570 | n->elems[j] = NULL; |
| 571 | n->kids[j+1] = NULL; |
| 572 | /* |
| 573 | * Now go round the deletion process again, with n |
| 574 | * pointing at the new big node and e still the same. |
| 575 | */ |
| 576 | n = a; |
| 577 | } |
| 578 | } |
| 579 | } |
| 580 | |
| 581 | /* |
| 582 | * Iterate over the elements of a tree234, in order. |
| 583 | */ |
| 584 | void *first234(tree234 *t, enum234 *e) { |
| 585 | node234 *n = t->root; |
| 586 | if (!n) |
| 587 | return NULL; |
| 588 | while (n->kids[0]) |
| 589 | n = n->kids[0]; |
| 590 | e->node = n; |
| 591 | e->posn = 0; |
| 592 | return n->elems[0]; |
| 593 | } |
| 594 | |
| 595 | void *next234(enum234 *e) { |
| 596 | node234 *n = e->node; |
| 597 | int pos = e->posn; |
| 598 | |
| 599 | if (n->kids[pos+1]) { |
| 600 | n = n->kids[pos+1]; |
| 601 | while (n->kids[0]) |
| 602 | n = n->kids[0]; |
| 603 | e->node = n; |
| 604 | e->posn = 0; |
| 605 | return n->elems[0]; |
| 606 | } |
| 607 | |
| 608 | if (pos < 2 && n->elems[pos+1]) { |
| 609 | e->posn = pos+1; |
| 610 | return n->elems[e->posn]; |
| 611 | } |
| 612 | |
| 613 | do { |
| 614 | node234 *nn = n->parent; |
| 615 | if (nn == NULL) |
| 616 | return NULL; /* end of tree */ |
| 617 | pos = (nn->kids[0] == n ? 0 : |
| 618 | nn->kids[1] == n ? 1 : |
| 619 | nn->kids[2] == n ? 2 : 3); |
| 620 | n = nn; |
| 621 | } while (pos == 3 || n->kids[pos+1] == NULL); |
| 622 | |
| 623 | e->node = n; |
| 624 | e->posn = pos; |
| 625 | return n->elems[pos]; |
| 626 | } |
| 627 | |
| 628 | #ifdef TEST |
| 629 | |
| 630 | int pnode(node234 *n, int level) { |
| 631 | printf("%*s%p\n", level*4, "", n); |
| 632 | if (n->kids[0]) pnode(n->kids[0], level+1); |
| 633 | if (n->elems[0]) printf("%*s\"%s\"\n", level*4+4, "", n->elems[0]); |
| 634 | if (n->kids[1]) pnode(n->kids[1], level+1); |
| 635 | if (n->elems[1]) printf("%*s\"%s\"\n", level*4+4, "", n->elems[1]); |
| 636 | if (n->kids[2]) pnode(n->kids[2], level+1); |
| 637 | if (n->elems[2]) printf("%*s\"%s\"\n", level*4+4, "", n->elems[2]); |
| 638 | if (n->kids[3]) pnode(n->kids[3], level+1); |
| 639 | } |
| 640 | int ptree(tree234 *t) { |
| 641 | if (t->root) |
| 642 | pnode(t->root, 0); |
| 643 | else |
| 644 | printf("empty tree\n"); |
| 645 | } |
| 646 | |
| 647 | int cmp(void *av, void *bv) { |
| 648 | char *a = (char *)av; |
| 649 | char *b = (char *)bv; |
| 650 | return strcmp(a, b); |
| 651 | } |
| 652 | |
| 653 | int main(void) { |
| 654 | tree234 *t = newtree234(cmp); |
| 655 | |
| 656 | add234(t, "Richard"); |
| 657 | add234(t, "Of"); |
| 658 | add234(t, "York"); |
| 659 | add234(t, "Gave"); |
| 660 | add234(t, "Battle"); |
| 661 | add234(t, "In"); |
| 662 | add234(t, "Vain"); |
| 663 | add234(t, "Rabbits"); |
| 664 | add234(t, "On"); |
| 665 | add234(t, "Your"); |
| 666 | add234(t, "Garden"); |
| 667 | add234(t, "Bring"); |
| 668 | add234(t, "Invisible"); |
| 669 | add234(t, "Vegetables"); |
| 670 | |
| 671 | ptree(t); |
| 672 | del234(t, find234(t, "Richard", NULL)); |
| 673 | ptree(t); |
| 674 | del234(t, find234(t, "Of", NULL)); |
| 675 | ptree(t); |
| 676 | del234(t, find234(t, "York", NULL)); |
| 677 | ptree(t); |
| 678 | del234(t, find234(t, "Gave", NULL)); |
| 679 | ptree(t); |
| 680 | del234(t, find234(t, "Battle", NULL)); |
| 681 | ptree(t); |
| 682 | del234(t, find234(t, "In", NULL)); |
| 683 | ptree(t); |
| 684 | del234(t, find234(t, "Vain", NULL)); |
| 685 | ptree(t); |
| 686 | del234(t, find234(t, "Rabbits", NULL)); |
| 687 | ptree(t); |
| 688 | del234(t, find234(t, "On", NULL)); |
| 689 | ptree(t); |
| 690 | del234(t, find234(t, "Your", NULL)); |
| 691 | ptree(t); |
| 692 | del234(t, find234(t, "Garden", NULL)); |
| 693 | ptree(t); |
| 694 | del234(t, find234(t, "Bring", NULL)); |
| 695 | ptree(t); |
| 696 | del234(t, find234(t, "Invisible", NULL)); |
| 697 | ptree(t); |
| 698 | del234(t, find234(t, "Vegetables", NULL)); |
| 699 | ptree(t); |
| 700 | } |
| 701 | #endif |