febd9a0f |
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) { |
81d77872 |
291 | if ( (c = cmp(e, n->elems[0])) < 0) |
febd9a0f |
292 | n = n->kids[0]; |
293 | else if (c == 0) |
294 | return n->elems[0]; |
81d77872 |
295 | else if (n->elems[1] == NULL || (c = cmp(e, n->elems[1])) < 0) |
febd9a0f |
296 | n = n->kids[1]; |
297 | else if (c == 0) |
298 | return n->elems[1]; |
81d77872 |
299 | else if (n->elems[2] == NULL || (c = cmp(e, n->elems[2])) < 0) |
febd9a0f |
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 | */ |
81d77872 |
319 | void del234(tree234 *t, void *e) { |
febd9a0f |
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 | n->elems[ki-1] = sib->elems[lastelem]; |
377 | sib->kids[lastelem+1] = NULL; |
378 | sib->elems[lastelem] = NULL; |
379 | LOG((" case 3a left\n")); |
380 | } else if (ki < 3 && n->kids[ki+1] && |
381 | n->kids[ki+1]->elems[1]) { |
382 | /* |
383 | * Case 3a, right-handed variant. ki has only |
384 | * one element but ki+1 has two or more. Move a |
385 | * subtree from ki+1 to ki. |
386 | * |
387 | * . B . . C . |
388 | * / \ -> / \ |
389 | * a A b c C d D e [more] a A b B c d D e [more] |
390 | */ |
391 | node234 *sib = n->kids[ki+1]; |
392 | int j; |
393 | sub->elems[1] = n->elems[ki]; |
394 | sub->kids[2] = sib->kids[0]; |
395 | n->elems[ki] = sib->elems[0]; |
396 | sib->kids[0] = sib->kids[1]; |
397 | for (j = 0; j < 2 && sib->elems[j+1]; j++) { |
398 | sib->kids[j+1] = sib->kids[j+2]; |
399 | sib->elems[j] = sib->elems[j+1]; |
400 | } |
401 | sib->kids[j+1] = NULL; |
402 | sib->elems[j] = NULL; |
403 | LOG((" case 3a right\n")); |
404 | } else { |
405 | /* |
406 | * Case 3b. ki has only one element, and has no |
407 | * neighbour with more than one. So pick a |
408 | * neighbour and merge it with ki, taking an |
409 | * element down from n to go in the middle. |
410 | * |
411 | * . B . . |
412 | * / \ -> | |
413 | * a A b c C d a A b B c C d |
414 | * |
415 | * (Since at all points we have avoided |
416 | * descending to a node with only one element, |
417 | * we can be sure that n is not reduced to |
418 | * nothingness by this move, _unless_ it was |
419 | * the very first node, ie the root of the |
420 | * tree. In that case we remove the now-empty |
421 | * root and replace it with its single large |
422 | * child as shown.) |
423 | */ |
424 | node234 *sib; |
425 | int j; |
426 | |
427 | if (ki > 0) |
428 | ki--; |
429 | sib = n->kids[ki]; |
430 | sub = n->kids[ki+1]; |
431 | |
432 | sub->kids[3] = sub->kids[1]; |
433 | sub->elems[2] = sub->elems[0]; |
434 | sub->kids[2] = sub->kids[0]; |
435 | sub->elems[1] = n->elems[ki]; |
436 | sub->kids[1] = sib->kids[1]; |
437 | sub->elems[0] = sib->elems[0]; |
438 | sub->kids[0] = sib->kids[0]; |
439 | |
440 | sfree(sib); |
441 | |
442 | /* |
443 | * That's built the big node in sub. Now we |
444 | * need to remove the reference to sib in n. |
445 | */ |
446 | for (j = ki; j < 3 && n->kids[j+1]; j++) { |
447 | n->kids[j] = n->kids[j+1]; |
448 | n->elems[j] = j<2 ? n->elems[j+1] : NULL; |
449 | } |
450 | n->kids[j] = NULL; |
451 | if (j < 3) n->elems[j] = NULL; |
452 | LOG((" case 3b\n")); |
453 | |
454 | if (!n->elems[0]) { |
455 | /* |
456 | * The root is empty and needs to be |
457 | * removed. |
458 | */ |
459 | LOG((" shifting root!\n")); |
460 | t->root = sub; |
461 | sub->parent = NULL; |
462 | sfree(n); |
463 | } |
464 | } |
465 | } |
466 | n = sub; |
467 | } |
468 | if (ei==-1) |
469 | return; /* nothing to do; `already removed' */ |
470 | |
471 | /* |
472 | * Treat special case: this is the one remaining item in |
473 | * the tree. n is the tree root (no parent), has one |
474 | * element (no elems[1]), and has no kids (no kids[0]). |
475 | */ |
476 | if (!n->parent && !n->elems[1] && !n->kids[0]) { |
477 | LOG((" removed last element in tree\n")); |
478 | sfree(n); |
479 | t->root = NULL; |
480 | return; |
481 | } |
482 | |
483 | /* |
484 | * Now we have the element we want, as n->elems[ei], and we |
485 | * have also arranged for that element not to be the only |
486 | * one in its node. So... |
487 | */ |
488 | |
489 | if (!n->kids[0] && n->elems[1]) { |
490 | /* |
491 | * Case 1. n is a leaf node with more than one element, |
492 | * so it's _really easy_. Just delete the thing and |
493 | * we're done. |
494 | */ |
495 | int i; |
496 | LOG((" case 1\n")); |
497 | for (i = ei; i < 3 && n->elems[i+1]; i++) |
498 | n->elems[i] = n->elems[i+1]; |
499 | n->elems[i] = NULL; |
500 | return; /* finished! */ |
501 | } else if (n->kids[ei]->elems[1]) { |
502 | /* |
503 | * Case 2a. n is an internal node, and the root of the |
504 | * subtree to the left of e has more than one element. |
505 | * So find the predecessor p to e (ie the largest node |
506 | * in that subtree), place it where e currently is, and |
507 | * then start the deletion process over again on the |
508 | * subtree with p as target. |
509 | */ |
510 | node234 *m = n->kids[ei]; |
511 | void *target; |
512 | LOG((" case 2a\n")); |
513 | while (m->kids[0]) { |
514 | m = (m->kids[3] ? m->kids[3] : |
515 | m->kids[2] ? m->kids[2] : |
516 | m->kids[1] ? m->kids[1] : m->kids[0]); |
517 | } |
518 | target = (m->elems[2] ? m->elems[2] : |
519 | m->elems[1] ? m->elems[1] : m->elems[0]); |
520 | n->elems[ei] = target; |
521 | n = n->kids[ei]; |
522 | e = target; |
523 | } else if (n->kids[ei+1]->elems[1]) { |
524 | /* |
525 | * Case 2b, symmetric to 2a but s/left/right/ and |
526 | * s/predecessor/successor/. (And s/largest/smallest/). |
527 | */ |
528 | node234 *m = n->kids[ei+1]; |
529 | void *target; |
530 | LOG((" case 2b\n")); |
531 | while (m->kids[0]) { |
532 | m = m->kids[0]; |
533 | } |
534 | target = m->elems[0]; |
535 | n->elems[ei] = target; |
536 | n = n->kids[ei+1]; |
537 | e = target; |
538 | } else { |
539 | /* |
540 | * Case 2c. n is an internal node, and the subtrees to |
541 | * the left and right of e both have only one element. |
542 | * So combine the two subnodes into a single big node |
543 | * with their own elements on the left and right and e |
544 | * in the middle, then restart the deletion process on |
545 | * that subtree, with e still as target. |
546 | */ |
547 | node234 *a = n->kids[ei], *b = n->kids[ei+1]; |
548 | int j; |
549 | |
550 | LOG((" case 2c\n")); |
551 | a->elems[1] = n->elems[ei]; |
552 | a->kids[2] = b->kids[0]; |
553 | a->elems[2] = b->elems[0]; |
554 | a->kids[3] = b->kids[1]; |
555 | sfree(b); |
556 | /* |
557 | * That's built the big node in a, and destroyed b. Now |
558 | * remove the reference to b (and e) in n. |
559 | */ |
560 | for (j = ei; j < 2 && n->elems[j+1]; j++) { |
561 | n->elems[j] = n->elems[j+1]; |
562 | n->kids[j+1] = n->kids[j+2]; |
563 | } |
564 | n->elems[j] = NULL; |
565 | n->kids[j+1] = NULL; |
566 | /* |
567 | * Now go round the deletion process again, with n |
568 | * pointing at the new big node and e still the same. |
569 | */ |
570 | n = a; |
571 | } |
572 | } |
573 | } |
574 | |
575 | /* |
576 | * Iterate over the elements of a tree234, in order. |
577 | */ |
578 | void *first234(tree234 *t, enum234 *e) { |
579 | node234 *n = t->root; |
580 | if (!n) |
581 | return NULL; |
582 | while (n->kids[0]) |
583 | n = n->kids[0]; |
584 | e->node = n; |
585 | e->posn = 0; |
586 | return n->elems[0]; |
587 | } |
588 | |
589 | void *next234(enum234 *e) { |
590 | node234 *n = e->node; |
591 | int pos = e->posn; |
592 | |
593 | if (n->kids[pos+1]) { |
594 | n = n->kids[pos+1]; |
595 | while (n->kids[0]) |
596 | n = n->kids[0]; |
597 | e->node = n; |
598 | e->posn = 0; |
599 | return n->elems[0]; |
600 | } |
601 | |
602 | if (pos == 0 && n->elems[1]) { |
603 | e->posn = 1; |
604 | return n->elems[1]; |
605 | } |
606 | |
607 | do { |
608 | node234 *nn = n->parent; |
609 | if (nn == NULL) |
610 | return NULL; /* end of tree */ |
611 | pos = (nn->kids[0] == n ? 0 : |
612 | nn->kids[1] == n ? 1 : |
613 | nn->kids[2] == n ? 2 : 3); |
614 | n = nn; |
615 | } while (pos == 3 || n->kids[pos+1] == NULL); |
616 | |
617 | e->node = n; |
618 | e->posn = pos; |
619 | return n->elems[pos]; |
620 | } |
621 | |
622 | #ifdef TEST |
623 | |
624 | int pnode(node234 *n, int level) { |
625 | printf("%*s%p\n", level*4, "", n); |
626 | if (n->kids[0]) pnode(n->kids[0], level+1); |
627 | if (n->elems[0]) printf("%*s\"%s\"\n", level*4+4, "", n->elems[0]); |
628 | if (n->kids[1]) pnode(n->kids[1], level+1); |
629 | if (n->elems[1]) printf("%*s\"%s\"\n", level*4+4, "", n->elems[1]); |
630 | if (n->kids[2]) pnode(n->kids[2], level+1); |
631 | if (n->elems[2]) printf("%*s\"%s\"\n", level*4+4, "", n->elems[2]); |
632 | if (n->kids[3]) pnode(n->kids[3], level+1); |
633 | } |
634 | int ptree(tree234 *t) { |
635 | if (t->root) |
636 | pnode(t->root, 0); |
637 | else |
638 | printf("empty tree\n"); |
639 | } |
640 | |
641 | int cmp(void *av, void *bv) { |
642 | char *a = (char *)av; |
643 | char *b = (char *)bv; |
644 | return strcmp(a, b); |
645 | } |
646 | |
647 | int main(void) { |
648 | tree234 *t = newtree234(cmp); |
649 | |
650 | add234(t, "Richard"); |
651 | add234(t, "Of"); |
652 | add234(t, "York"); |
653 | add234(t, "Gave"); |
654 | add234(t, "Battle"); |
655 | add234(t, "In"); |
656 | add234(t, "Vain"); |
657 | add234(t, "Rabbits"); |
658 | add234(t, "On"); |
659 | add234(t, "Your"); |
660 | add234(t, "Garden"); |
661 | add234(t, "Bring"); |
662 | add234(t, "Invisible"); |
663 | add234(t, "Vegetables"); |
664 | |
665 | ptree(t); |
666 | del234(t, find234(t, "Richard", NULL)); |
667 | ptree(t); |
668 | del234(t, find234(t, "Of", NULL)); |
669 | ptree(t); |
670 | del234(t, find234(t, "York", NULL)); |
671 | ptree(t); |
672 | del234(t, find234(t, "Gave", NULL)); |
673 | ptree(t); |
674 | del234(t, find234(t, "Battle", NULL)); |
675 | ptree(t); |
676 | del234(t, find234(t, "In", NULL)); |
677 | ptree(t); |
678 | del234(t, find234(t, "Vain", NULL)); |
679 | ptree(t); |
680 | del234(t, find234(t, "Rabbits", NULL)); |
681 | ptree(t); |
682 | del234(t, find234(t, "On", NULL)); |
683 | ptree(t); |
684 | del234(t, find234(t, "Your", NULL)); |
685 | ptree(t); |
686 | del234(t, find234(t, "Garden", NULL)); |
687 | ptree(t); |
688 | del234(t, find234(t, "Bring", NULL)); |
689 | ptree(t); |
690 | del234(t, find234(t, "Invisible", NULL)); |
691 | ptree(t); |
692 | del234(t, find234(t, "Vegetables", NULL)); |
693 | ptree(t); |
694 | } |
695 | #endif |