1 /* Copyright (C) 1995-1997, 2000, 2006 Free Software Foundation, Inc.
2    Contributed by Bernd Schmidt <crux@Pool.Informatik.RWTH-Aachen.DE>, 1997.
3 
4    NOTE: The canonical source of this file is maintained with the GNU C
5    Library.  Bugs can be reported to bug-glibc@gnu.org.
6 
7    This program is free software: you can redistribute it and/or modify
8    it under the terms of the GNU Lesser General Public License as published by
9    the Free Software Foundation; either version 2.1 of the License, or
10    (at your option) any later version.
11 
12    This program is distributed in the hope that it will be useful,
13    but WITHOUT ANY WARRANTY; without even the implied warranty of
14    MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
15    GNU Lesser General Public License for more details.
16 
17    You should have received a copy of the GNU Lesser General Public License
18    along with this program.  If not, see <https://www.gnu.org/licenses/>.  */
19 
20 /* Tree search for red/black trees.
21    The algorithm for adding nodes is taken from one of the many "Algorithms"
22    books by Robert Sedgewick, although the implementation differs.
23    The algorithm for deleting nodes can probably be found in a book named
24    "Introduction to Algorithms" by Cormen/Leiserson/Rivest.  At least that's
25    the book that my professor took most algorithms from during the "Data
26    Structures" course...
27 
28    Totally public domain.  */
29 
30 /* Red/black trees are binary trees in which the edges are colored either red
31    or black.  They have the following properties:
32    1. The number of black edges on every path from the root to a leaf is
33       constant.
34    2. No two red edges are adjacent.
35    Therefore there is an upper bound on the length of every path, it's
36    O(log n) where n is the number of nodes in the tree.  No path can be longer
37    than 1+2*P where P is the length of the shortest path in the tree.
38    Useful for the implementation:
39    3. If one of the children of a node is NULL, then the other one is red
40       (if it exists).
41 
42    In the implementation, not the edges are colored, but the nodes.  The color
43    interpreted as the color of the edge leading to this node.  The color is
44    meaningless for the root node, but we color the root node black for
45    convenience.  All added nodes are red initially.
46 
47    Adding to a red/black tree is rather easy.  The right place is searched
48    with a usual binary tree search.  Additionally, whenever a node N is
49    reached that has two red successors, the successors are colored black and
50    the node itself colored red.  This moves red edges up the tree where they
51    pose less of a problem once we get to really insert the new node.  Changing
52    N's color to red may violate rule 2, however, so rotations may become
53    necessary to restore the invariants.  Adding a new red leaf may violate
54    the same rule, so afterwards an additional check is run and the tree
55    possibly rotated.
56 
57    Deleting is hairy.  There are mainly two nodes involved: the node to be
58    deleted (n1), and another node that is to be unchained from the tree (n2).
59    If n1 has a successor (the node with a smallest key that is larger than
60    n1), then the successor becomes n2 and its contents are copied into n1,
61    otherwise n1 becomes n2.
62    Unchaining a node may violate rule 1: if n2 is black, one subtree is
63    missing one black edge afterwards.  The algorithm must try to move this
64    error upwards towards the root, so that the subtree that does not have
65    enough black edges becomes the whole tree.  Once that happens, the error
66    has disappeared.  It may not be necessary to go all the way up, since it
67    is possible that rotations and recoloring can fix the error before that.
68 
69    Although the deletion algorithm must walk upwards through the tree, we
70    do not store parent pointers in the nodes.  Instead, delete allocates a
71    small array of parent pointers and fills it while descending the tree.
72    Since we know that the length of a path is O(log n), where n is the number
73    of nodes, this is likely to use less memory.  */
74 
75 /* Tree rotations look like this:
76       A                C
77      / \              / \
78     B   C            A   G
79    / \ / \  -->     / \
80    D E F G         B   F
81                   / \
82                  D   E
83 
84    In this case, A has been rotated left.  This preserves the ordering of the
85    binary tree.  */
86 
87 #include <config.h>
88 
89 /* Specification.  */
90 #ifdef IN_LIBINTL
91 # include "tsearch.h"
92 #else
93 # include <search.h>
94 #endif
95 
96 #include <stdlib.h>
97 
98 typedef int (*__compar_fn_t) (const void *, const void *);
99 typedef void (*__action_fn_t) (const void *, VISIT, int);
100 
101 #ifndef weak_alias
102 # define __tsearch tsearch
103 # define __tfind tfind
104 # define __tdelete tdelete
105 # define __twalk twalk
106 #endif
107 
108 #ifndef internal_function
109 /* Inside GNU libc we mark some function in a special way.  In other
110    environments simply ignore the marking.  */
111 # define internal_function
112 #endif
113 
114 typedef struct node_t
115 {
116   /* Callers expect this to be the first element in the structure - do not
117      move!  */
118   const void *key;
119   struct node_t *left;
120   struct node_t *right;
121   unsigned int red:1;
122 } *node;
123 typedef const struct node_t *const_node;
124 
125 #undef DEBUGGING
126 
127 #ifdef DEBUGGING
128 
129 /* Routines to check tree invariants.  */
130 
131 #include <assert.h>
132 
133 #define CHECK_TREE(a) check_tree(a)
134 
135 static void
check_tree_recurse(node p,int d_sofar,int d_total)136 check_tree_recurse (node p, int d_sofar, int d_total)
137 {
138   if (p == NULL)
139     {
140       assert (d_sofar == d_total);
141       return;
142     }
143 
144   check_tree_recurse (p->left, d_sofar + (p->left && !p->left->red), d_total);
145   check_tree_recurse (p->right, d_sofar + (p->right && !p->right->red), d_total);
146   if (p->left)
147     assert (!(p->left->red && p->red));
148   if (p->right)
149     assert (!(p->right->red && p->red));
150 }
151 
152 static void
check_tree(node root)153 check_tree (node root)
154 {
155   int cnt = 0;
156   node p;
157   if (root == NULL)
158     return;
159   root->red = 0;
160   for(p = root->left; p; p = p->left)
161     cnt += !p->red;
162   check_tree_recurse (root, 0, cnt);
163 }
164 
165 
166 #else
167 
168 #define CHECK_TREE(a)
169 
170 #endif
171 
172 /* Possibly "split" a node with two red successors, and/or fix up two red
173    edges in a row.  ROOTP is a pointer to the lowest node we visited, PARENTP
174    and GPARENTP pointers to its parent/grandparent.  P_R and GP_R contain the
175    comparison values that determined which way was taken in the tree to reach
176    ROOTP.  MODE is 1 if we need not do the split, but must check for two red
177    edges between GPARENTP and ROOTP.  */
178 static void
maybe_split_for_insert(node * rootp,node * parentp,node * gparentp,int p_r,int gp_r,int mode)179 maybe_split_for_insert (node *rootp, node *parentp, node *gparentp,
180                         int p_r, int gp_r, int mode)
181 {
182   node root = *rootp;
183   node *rp, *lp;
184   rp = &(*rootp)->right;
185   lp = &(*rootp)->left;
186 
187   /* See if we have to split this node (both successors red).  */
188   if (mode == 1
189       || ((*rp) != NULL && (*lp) != NULL && (*rp)->red && (*lp)->red))
190     {
191       /* This node becomes red, its successors black.  */
192       root->red = 1;
193       if (*rp)
194         (*rp)->red = 0;
195       if (*lp)
196         (*lp)->red = 0;
197 
198       /* If the parent of this node is also red, we have to do
199          rotations.  */
200       if (parentp != NULL && (*parentp)->red)
201         {
202           node gp = *gparentp;
203           node p = *parentp;
204           /* There are two main cases:
205              1. The edge types (left or right) of the two red edges differ.
206              2. Both red edges are of the same type.
207              There exist two symmetries of each case, so there is a total of
208              4 cases.  */
209           if ((p_r > 0) != (gp_r > 0))
210             {
211               /* Put the child at the top of the tree, with its parent
212                  and grandparent as successors.  */
213               p->red = 1;
214               gp->red = 1;
215               root->red = 0;
216               if (p_r < 0)
217                 {
218                   /* Child is left of parent.  */
219                   p->left = *rp;
220                   *rp = p;
221                   gp->right = *lp;
222                   *lp = gp;
223                 }
224               else
225                 {
226                   /* Child is right of parent.  */
227                   p->right = *lp;
228                   *lp = p;
229                   gp->left = *rp;
230                   *rp = gp;
231                 }
232               *gparentp = root;
233             }
234           else
235             {
236               *gparentp = *parentp;
237               /* Parent becomes the top of the tree, grandparent and
238                  child are its successors.  */
239               p->red = 0;
240               gp->red = 1;
241               if (p_r < 0)
242                 {
243                   /* Left edges.  */
244                   gp->left = p->right;
245                   p->right = gp;
246                 }
247               else
248                 {
249                   /* Right edges.  */
250                   gp->right = p->left;
251                   p->left = gp;
252                 }
253             }
254         }
255     }
256 }
257 
258 /* Find or insert datum into search tree.
259    KEY is the key to be located, ROOTP is the address of tree root,
260    COMPAR the ordering function.  */
261 void *
__tsearch(const void * key,void ** vrootp,__compar_fn_t compar)262 __tsearch (const void *key, void **vrootp, __compar_fn_t compar)
263 {
264   node q;
265   node *parentp = NULL, *gparentp = NULL;
266   node *rootp = (node *) vrootp;
267   node *nextp;
268   int r = 0, p_r = 0, gp_r = 0; /* No they might not, Mr Compiler.  */
269 
270   if (rootp == NULL)
271     return NULL;
272 
273   /* This saves some additional tests below.  */
274   if (*rootp != NULL)
275     (*rootp)->red = 0;
276 
277   CHECK_TREE (*rootp);
278 
279   nextp = rootp;
280   while (*nextp != NULL)
281     {
282       node root = *rootp;
283       r = (*compar) (key, root->key);
284       if (r == 0)
285         return root;
286 
287       maybe_split_for_insert (rootp, parentp, gparentp, p_r, gp_r, 0);
288       /* If that did any rotations, parentp and gparentp are now garbage.
289          That doesn't matter, because the values they contain are never
290          used again in that case.  */
291 
292       nextp = r < 0 ? &root->left : &root->right;
293       if (*nextp == NULL)
294         break;
295 
296       gparentp = parentp;
297       parentp = rootp;
298       rootp = nextp;
299 
300       gp_r = p_r;
301       p_r = r;
302     }
303 
304   q = (struct node_t *) malloc (sizeof (struct node_t));
305   if (q != NULL)
306     {
307       *nextp = q;                       /* link new node to old */
308       q->key = key;                     /* initialize new node */
309       q->red = 1;
310       q->left = q->right = NULL;
311 
312       if (nextp != rootp)
313         /* There may be two red edges in a row now, which we must avoid by
314            rotating the tree.  */
315         maybe_split_for_insert (nextp, rootp, parentp, r, p_r, 1);
316     }
317 
318   return q;
319 }
320 #ifdef weak_alias
321 weak_alias (__tsearch, tsearch)
322 #endif
323 
324 
325 /* Find datum in search tree.
326    KEY is the key to be located, ROOTP is the address of tree root,
327    COMPAR the ordering function.  */
328 void *
329 __tfind (key, vrootp, compar)
330      const void *key;
331      void *const *vrootp;
332      __compar_fn_t compar;
333 {
334   node *rootp = (node *) vrootp;
335 
336   if (rootp == NULL)
337     return NULL;
338 
339   CHECK_TREE (*rootp);
340 
341   while (*rootp != NULL)
342     {
343       node root = *rootp;
344       int r;
345 
346       r = (*compar) (key, root->key);
347       if (r == 0)
348         return root;
349 
350       rootp = r < 0 ? &root->left : &root->right;
351     }
352   return NULL;
353 }
354 #ifdef weak_alias
weak_alias(__tfind,tfind)355 weak_alias (__tfind, tfind)
356 #endif
357 
358 
359 /* Delete node with given key.
360    KEY is the key to be deleted, ROOTP is the address of the root of tree,
361    COMPAR the comparison function.  */
362 void *
363 __tdelete (const void *key, void **vrootp, __compar_fn_t compar)
364 {
365   node p, q, r, retval;
366   int cmp;
367   node *rootp = (node *) vrootp;
368   node root, unchained;
369   /* Stack of nodes so we remember the parents without recursion.  It's
370      _very_ unlikely that there are paths longer than 40 nodes.  The tree
371      would need to have around 250.000 nodes.  */
372   int stacksize = 100;
373   int sp = 0;
374   node *nodestack[100];
375 
376   if (rootp == NULL)
377     return NULL;
378   p = *rootp;
379   if (p == NULL)
380     return NULL;
381 
382   CHECK_TREE (p);
383 
384   while ((cmp = (*compar) (key, (*rootp)->key)) != 0)
385     {
386       if (sp == stacksize)
387         abort ();
388 
389       nodestack[sp++] = rootp;
390       p = *rootp;
391       rootp = ((cmp < 0)
392                ? &(*rootp)->left
393                : &(*rootp)->right);
394       if (*rootp == NULL)
395         return NULL;
396     }
397 
398   /* This is bogus if the node to be deleted is the root... this routine
399      really should return an integer with 0 for success, -1 for failure
400      and errno = ESRCH or something.  */
401   retval = p;
402 
403   /* We don't unchain the node we want to delete. Instead, we overwrite
404      it with its successor and unchain the successor.  If there is no
405      successor, we really unchain the node to be deleted.  */
406 
407   root = *rootp;
408 
409   r = root->right;
410   q = root->left;
411 
412   if (q == NULL || r == NULL)
413     unchained = root;
414   else
415     {
416       node *parent = rootp, *up = &root->right;
417       for (;;)
418         {
419           if (sp == stacksize)
420             abort ();
421           nodestack[sp++] = parent;
422           parent = up;
423           if ((*up)->left == NULL)
424             break;
425           up = &(*up)->left;
426         }
427       unchained = *up;
428     }
429 
430   /* We know that either the left or right successor of UNCHAINED is NULL.
431      R becomes the other one, it is chained into the parent of UNCHAINED.  */
432   r = unchained->left;
433   if (r == NULL)
434     r = unchained->right;
435   if (sp == 0)
436     *rootp = r;
437   else
438     {
439       q = *nodestack[sp-1];
440       if (unchained == q->right)
441         q->right = r;
442       else
443         q->left = r;
444     }
445 
446   if (unchained != root)
447     root->key = unchained->key;
448   if (!unchained->red)
449     {
450       /* Now we lost a black edge, which means that the number of black
451          edges on every path is no longer constant.  We must balance the
452          tree.  */
453       /* NODESTACK now contains all parents of R.  R is likely to be NULL
454          in the first iteration.  */
455       /* NULL nodes are considered black throughout - this is necessary for
456          correctness.  */
457       while (sp > 0 && (r == NULL || !r->red))
458         {
459           node *pp = nodestack[sp - 1];
460           p = *pp;
461           /* Two symmetric cases.  */
462           if (r == p->left)
463             {
464               /* Q is R's brother, P is R's parent.  The subtree with root
465                  R has one black edge less than the subtree with root Q.  */
466               q = p->right;
467               if (q->red)
468                 {
469                   /* If Q is red, we know that P is black. We rotate P left
470                      so that Q becomes the top node in the tree, with P below
471                      it.  P is colored red, Q is colored black.
472                      This action does not change the black edge count for any
473                      leaf in the tree, but we will be able to recognize one
474                      of the following situations, which all require that Q
475                      is black.  */
476                   q->red = 0;
477                   p->red = 1;
478                   /* Left rotate p.  */
479                   p->right = q->left;
480                   q->left = p;
481                   *pp = q;
482                   /* Make sure pp is right if the case below tries to use
483                      it.  */
484                   nodestack[sp++] = pp = &q->left;
485                   q = p->right;
486                 }
487               /* We know that Q can't be NULL here.  We also know that Q is
488                  black.  */
489               if ((q->left == NULL || !q->left->red)
490                   && (q->right == NULL || !q->right->red))
491                 {
492                   /* Q has two black successors.  We can simply color Q red.
493                      The whole subtree with root P is now missing one black
494                      edge.  Note that this action can temporarily make the
495                      tree invalid (if P is red).  But we will exit the loop
496                      in that case and set P black, which both makes the tree
497                      valid and also makes the black edge count come out
498                      right.  If P is black, we are at least one step closer
499                      to the root and we'll try again the next iteration.  */
500                   q->red = 1;
501                   r = p;
502                 }
503               else
504                 {
505                   /* Q is black, one of Q's successors is red.  We can
506                      repair the tree with one operation and will exit the
507                      loop afterwards.  */
508                   if (q->right == NULL || !q->right->red)
509                     {
510                       /* The left one is red.  We perform the same action as
511                          in maybe_split_for_insert where two red edges are
512                          adjacent but point in different directions:
513                          Q's left successor (let's call it Q2) becomes the
514                          top of the subtree we are looking at, its parent (Q)
515                          and grandparent (P) become its successors. The former
516                          successors of Q2 are placed below P and Q.
517                          P becomes black, and Q2 gets the color that P had.
518                          This changes the black edge count only for node R and
519                          its successors.  */
520                       node q2 = q->left;
521                       q2->red = p->red;
522                       p->right = q2->left;
523                       q->left = q2->right;
524                       q2->right = q;
525                       q2->left = p;
526                       *pp = q2;
527                       p->red = 0;
528                     }
529                   else
530                     {
531                       /* It's the right one.  Rotate P left. P becomes black,
532                          and Q gets the color that P had.  Q's right successor
533                          also becomes black.  This changes the black edge
534                          count only for node R and its successors.  */
535                       q->red = p->red;
536                       p->red = 0;
537 
538                       q->right->red = 0;
539 
540                       /* left rotate p */
541                       p->right = q->left;
542                       q->left = p;
543                       *pp = q;
544                     }
545 
546                   /* We're done.  */
547                   sp = 1;
548                   r = NULL;
549                 }
550             }
551           else
552             {
553               /* Comments: see above.  */
554               q = p->left;
555               if (q->red)
556                 {
557                   q->red = 0;
558                   p->red = 1;
559                   p->left = q->right;
560                   q->right = p;
561                   *pp = q;
562                   nodestack[sp++] = pp = &q->right;
563                   q = p->left;
564                 }
565               if ((q->right == NULL || !q->right->red)
566                        && (q->left == NULL || !q->left->red))
567                 {
568                   q->red = 1;
569                   r = p;
570                 }
571               else
572                 {
573                   if (q->left == NULL || !q->left->red)
574                     {
575                       node q2 = q->right;
576                       q2->red = p->red;
577                       p->left = q2->right;
578                       q->right = q2->left;
579                       q2->left = q;
580                       q2->right = p;
581                       *pp = q2;
582                       p->red = 0;
583                     }
584                   else
585                     {
586                       q->red = p->red;
587                       p->red = 0;
588                       q->left->red = 0;
589                       p->left = q->right;
590                       q->right = p;
591                       *pp = q;
592                     }
593                   sp = 1;
594                   r = NULL;
595                 }
596             }
597           --sp;
598         }
599       if (r != NULL)
600         r->red = 0;
601     }
602 
603   free (unchained);
604   return retval;
605 }
606 #ifdef weak_alias
weak_alias(__tdelete,tdelete)607 weak_alias (__tdelete, tdelete)
608 #endif
609 
610 
611 /* Walk the nodes of a tree.
612    ROOT is the root of the tree to be walked, ACTION the function to be
613    called at each node.  LEVEL is the level of ROOT in the whole tree.  */
614 static void
615 internal_function
616 trecurse (const void *vroot, __action_fn_t action, int level)
617 {
618   const_node root = (const_node) vroot;
619 
620   if (root->left == NULL && root->right == NULL)
621     (*action) (root, leaf, level);
622   else
623     {
624       (*action) (root, preorder, level);
625       if (root->left != NULL)
626         trecurse (root->left, action, level + 1);
627       (*action) (root, postorder, level);
628       if (root->right != NULL)
629         trecurse (root->right, action, level + 1);
630       (*action) (root, endorder, level);
631     }
632 }
633 
634 
635 /* Walk the nodes of a tree.
636    ROOT is the root of the tree to be walked, ACTION the function to be
637    called at each node.  */
638 void
__twalk(const void * vroot,__action_fn_t action)639 __twalk (const void *vroot, __action_fn_t action)
640 {
641   const_node root = (const_node) vroot;
642 
643   CHECK_TREE (root);
644 
645   if (root != NULL && action != NULL)
646     trecurse (root, action, 0);
647 }
648 #ifdef weak_alias
weak_alias(__twalk,twalk)649 weak_alias (__twalk, twalk)
650 #endif
651 
652 
653 #ifdef _LIBC
654 
655 /* The standardized functions miss an important functionality: the
656    tree cannot be removed easily.  We provide a function to do this.  */
657 static void
658 internal_function
659 tdestroy_recurse (node root, __free_fn_t freefct)
660 {
661   if (root->left != NULL)
662     tdestroy_recurse (root->left, freefct);
663   if (root->right != NULL)
664     tdestroy_recurse (root->right, freefct);
665   (*freefct) ((void *) root->key);
666   /* Free the node itself.  */
667   free (root);
668 }
669 
670 void
__tdestroy(void * vroot,__free_fn_t freefct)671 __tdestroy (void *vroot, __free_fn_t freefct)
672 {
673   node root = (node) vroot;
674 
675   CHECK_TREE (root);
676 
677   if (root != NULL)
678     tdestroy_recurse (root, freefct);
679 }
680 weak_alias (__tdestroy, tdestroy)
681 
682 #endif /* _LIBC */
683