Lines Matching refs:right
49 Adding to a red/black tree is rather easy. The right place is searched
122 struct node_t *right;
147 check_tree_recurse (p->right, d_sofar + (p->right && !p->right->red), d_total);
150 if (p->right)
151 assert (!(p->right->red && p->red));
186 rp = &(*rootp)->right;
207 1. The edge types (left or right) of the two red edges differ.
223 gp->right = *lp;
228 /* Child is right of parent. */
229 p->right = *lp;
246 gp->left = p->right;
247 p->right = gp;
252 gp->right = p->left;
294 nextp = r < 0 ? &root->left : &root->right;
312 q->left = q->right = NULL;
352 rootp = r < 0 ? &root->left : &root->right;
395 : &(*rootp)->right);
411 r = root->right;
418 node *parent = rootp, *up = &root->right;
432 /* We know that either the left or right successor of UNCHAINED is NULL.
436 r = unchained->right;
442 if (unchained == q->right)
443 q->right = r;
468 q = p->right;
481 p->right = q->left;
484 /* Make sure pp is right if the case below tries to use
487 q = p->right;
492 && (q->right == NULL || !q->right->red))
500 right. If P is black, we are at least one step closer
510 if (q->right == NULL || !q->right->red)
524 p->right = q2->left;
525 q->left = q2->right;
526 q2->right = q;
533 /* It's the right one. Rotate P left. P becomes black,
534 and Q gets the color that P had. Q's right successor
540 q->right->red = 0;
543 p->right = q->left;
561 p->left = q->right;
562 q->right = p;
564 nodestack[sp++] = pp = &q->right;
567 if ((q->right == NULL || !q->right->red)
577 node q2 = q->right;
579 p->left = q2->right;
580 q->right = q2->left;
582 q2->right = p;
591 p->left = q->right;
592 q->right = p;
622 if (root->left == NULL && root->right == NULL)
630 if (root->right != NULL)
631 trecurse (root->right, action, level + 1);
665 if (root->right != NULL)
666 tdestroy_recurse (root->right, freefct);