mirror of https://github.com/aria2/aria2
				
				
				
			
		
			
				
	
	
		
			685 lines
		
	
	
		
			18 KiB
		
	
	
	
		
			C
		
	
	
			
		
		
	
	
			685 lines
		
	
	
		
			18 KiB
		
	
	
	
		
			C
		
	
	
/* Copyright (C) 1995, 1996, 1997, 2000, 2006 Free Software Foundation, Inc.
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   Contributed by Bernd Schmidt <crux@Pool.Informatik.RWTH-Aachen.DE>, 1997.
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   NOTE: The canonical source of this file is maintained with the GNU C
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   Library.  Bugs can be reported to bug-glibc@gnu.org.
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   This program is free software; you can redistribute it and/or modify it
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   under the terms of the GNU Library General Public License as published
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   by the Free Software Foundation; either version 2, or (at your option)
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   any later version.
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   This program is distributed in the hope that it will be useful,
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   but WITHOUT ANY WARRANTY; without even the implied warranty of
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   MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
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   Library General Public License for more details.
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   You should have received a copy of the GNU Library General Public
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   License along with this program; if not, write to the Free Software
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   Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301,
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   USA.  */
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/* Tree search for red/black trees.
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   The algorithm for adding nodes is taken from one of the many "Algorithms"
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   books by Robert Sedgewick, although the implementation differs.
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   The algorithm for deleting nodes can probably be found in a book named
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   "Introduction to Algorithms" by Cormen/Leiserson/Rivest.  At least that's
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   the book that my professor took most algorithms from during the "Data
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   Structures" course...
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   Totally public domain.  */
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/* Red/black trees are binary trees in which the edges are colored either red
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   or black.  They have the following properties:
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   1. The number of black edges on every path from the root to a leaf is
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      constant.
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   2. No two red edges are adjacent.
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   Therefore there is an upper bound on the length of every path, it's
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   O(log n) where n is the number of nodes in the tree.  No path can be longer
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   than 1+2*P where P is the length of the shortest path in the tree.
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   Useful for the implementation:
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   3. If one of the children of a node is NULL, then the other one is red
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      (if it exists).
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   In the implementation, not the edges are colored, but the nodes.  The color
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   interpreted as the color of the edge leading to this node.  The color is
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   meaningless for the root node, but we color the root node black for
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   convenience.  All added nodes are red initially.
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   Adding to a red/black tree is rather easy.  The right place is searched
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   with a usual binary tree search.  Additionally, whenever a node N is
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   reached that has two red successors, the successors are colored black and
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   the node itself colored red.  This moves red edges up the tree where they
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   pose less of a problem once we get to really insert the new node.  Changing
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   N's color to red may violate rule 2, however, so rotations may become
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   necessary to restore the invariants.  Adding a new red leaf may violate
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   the same rule, so afterwards an additional check is run and the tree
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   possibly rotated.
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   Deleting is hairy.  There are mainly two nodes involved: the node to be
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   deleted (n1), and another node that is to be unchained from the tree (n2).
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   If n1 has a successor (the node with a smallest key that is larger than
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   n1), then the successor becomes n2 and its contents are copied into n1,
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   otherwise n1 becomes n2.
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   Unchaining a node may violate rule 1: if n2 is black, one subtree is
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   missing one black edge afterwards.  The algorithm must try to move this
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   error upwards towards the root, so that the subtree that does not have
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   enough black edges becomes the whole tree.  Once that happens, the error
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   has disappeared.  It may not be necessary to go all the way up, since it
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   is possible that rotations and recoloring can fix the error before that.
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   Although the deletion algorithm must walk upwards through the tree, we
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   do not store parent pointers in the nodes.  Instead, delete allocates a
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   small array of parent pointers and fills it while descending the tree.
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   Since we know that the length of a path is O(log n), where n is the number
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   of nodes, this is likely to use less memory.  */
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/* Tree rotations look like this:
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      A                C
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     / \              / \
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    B   C            A   G
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   / \ / \  -->     / \
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   D E F G         B   F
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                  / \
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                 D   E
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   In this case, A has been rotated left.  This preserves the ordering of the
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   binary tree.  */
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#include <config.h>
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/* Specification.  */
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#ifdef IN_LIBINTL
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# include "tsearch.h"
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#else
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# include <search.h>
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#endif
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#include <stdlib.h>
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typedef int (*__compar_fn_t) (const void *, const void *);
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typedef void (*__action_fn_t) (const void *, VISIT, int);
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#ifndef weak_alias
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# define __tsearch tsearch
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# define __tfind tfind
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# define __tdelete tdelete
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# define __twalk twalk
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#endif
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#ifndef internal_function
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/* Inside GNU libc we mark some function in a special way.  In other
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   environments simply ignore the marking.  */
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# define internal_function
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#endif
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typedef struct node_t
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{
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  /* Callers expect this to be the first element in the structure - do not
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     move!  */
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  const void *key;
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  struct node_t *left;
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  struct node_t *right;
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  unsigned int red:1;
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} *node;
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typedef const struct node_t *const_node;
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#undef DEBUGGING
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#ifdef DEBUGGING
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/* Routines to check tree invariants.  */
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#include <assert.h>
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#define CHECK_TREE(a) check_tree(a)
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static void
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check_tree_recurse (node p, int d_sofar, int d_total)
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{
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  if (p == NULL)
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    {
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      assert (d_sofar == d_total);
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      return;
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    }
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  check_tree_recurse (p->left, d_sofar + (p->left && !p->left->red), d_total);
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  check_tree_recurse (p->right, d_sofar + (p->right && !p->right->red), d_total);
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  if (p->left)
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    assert (!(p->left->red && p->red));
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  if (p->right)
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    assert (!(p->right->red && p->red));
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}
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static void
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check_tree (node root)
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{
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  int cnt = 0;
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  node p;
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  if (root == NULL)
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    return;
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  root->red = 0;
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  for(p = root->left; p; p = p->left)
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    cnt += !p->red;
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  check_tree_recurse (root, 0, cnt);
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}
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#else
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#define CHECK_TREE(a)
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#endif
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/* Possibly "split" a node with two red successors, and/or fix up two red
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   edges in a row.  ROOTP is a pointer to the lowest node we visited, PARENTP
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   and GPARENTP pointers to its parent/grandparent.  P_R and GP_R contain the
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   comparison values that determined which way was taken in the tree to reach
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   ROOTP.  MODE is 1 if we need not do the split, but must check for two red
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   edges between GPARENTP and ROOTP.  */
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static void
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maybe_split_for_insert (node *rootp, node *parentp, node *gparentp,
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			int p_r, int gp_r, int mode)
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{
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  node root = *rootp;
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  node *rp, *lp;
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  rp = &(*rootp)->right;
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  lp = &(*rootp)->left;
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  /* See if we have to split this node (both successors red).  */
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  if (mode == 1
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      || ((*rp) != NULL && (*lp) != NULL && (*rp)->red && (*lp)->red))
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    {
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      /* This node becomes red, its successors black.  */
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      root->red = 1;
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      if (*rp)
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	(*rp)->red = 0;
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      if (*lp)
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	(*lp)->red = 0;
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      /* If the parent of this node is also red, we have to do
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	 rotations.  */
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      if (parentp != NULL && (*parentp)->red)
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	{
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	  node gp = *gparentp;
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	  node p = *parentp;
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	  /* There are two main cases:
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	     1. The edge types (left or right) of the two red edges differ.
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	     2. Both red edges are of the same type.
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	     There exist two symmetries of each case, so there is a total of
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	     4 cases.  */
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	  if ((p_r > 0) != (gp_r > 0))
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	    {
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	      /* Put the child at the top of the tree, with its parent
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		 and grandparent as successors.  */
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	      p->red = 1;
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	      gp->red = 1;
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	      root->red = 0;
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	      if (p_r < 0)
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		{
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		  /* Child is left of parent.  */
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		  p->left = *rp;
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		  *rp = p;
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		  gp->right = *lp;
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		  *lp = gp;
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		}
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	      else
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		{
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		  /* Child is right of parent.  */
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		  p->right = *lp;
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		  *lp = p;
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		  gp->left = *rp;
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		  *rp = gp;
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		}
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	      *gparentp = root;
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	    }
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	  else
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	    {
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	      *gparentp = *parentp;
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	      /* Parent becomes the top of the tree, grandparent and
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		 child are its successors.  */
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	      p->red = 0;
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	      gp->red = 1;
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	      if (p_r < 0)
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		{
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		  /* Left edges.  */
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		  gp->left = p->right;
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		  p->right = gp;
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		}
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	      else
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		{
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		  /* Right edges.  */
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		  gp->right = p->left;
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		  p->left = gp;
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		}
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	    }
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	}
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    }
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}
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/* Find or insert datum into search tree.
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   KEY is the key to be located, ROOTP is the address of tree root,
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   COMPAR the ordering function.  */
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void *
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__tsearch (const void *key, void **vrootp, __compar_fn_t compar)
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{
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  node q;
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  node *parentp = NULL, *gparentp = NULL;
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  node *rootp = (node *) vrootp;
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  node *nextp;
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  int r = 0, p_r = 0, gp_r = 0; /* No they might not, Mr Compiler.  */
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  if (rootp == NULL)
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    return NULL;
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  /* This saves some additional tests below.  */
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  if (*rootp != NULL)
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    (*rootp)->red = 0;
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  CHECK_TREE (*rootp);
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  nextp = rootp;
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  while (*nextp != NULL)
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    {
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      node root = *rootp;
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      r = (*compar) (key, root->key);
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      if (r == 0)
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	return root;
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      maybe_split_for_insert (rootp, parentp, gparentp, p_r, gp_r, 0);
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      /* If that did any rotations, parentp and gparentp are now garbage.
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	 That doesn't matter, because the values they contain are never
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	 used again in that case.  */
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      nextp = r < 0 ? &root->left : &root->right;
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      if (*nextp == NULL)
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	break;
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      gparentp = parentp;
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      parentp = rootp;
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      rootp = nextp;
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      gp_r = p_r;
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      p_r = r;
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    }
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  q = (struct node_t *) malloc (sizeof (struct node_t));
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  if (q != NULL)
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    {
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      *nextp = q;			/* link new node to old */
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      q->key = key;			/* initialize new node */
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      q->red = 1;
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      q->left = q->right = NULL;
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      if (nextp != rootp)
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	/* There may be two red edges in a row now, which we must avoid by
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	   rotating the tree.  */
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	maybe_split_for_insert (nextp, rootp, parentp, r, p_r, 1);
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    }
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  return q;
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}
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#ifdef weak_alias
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weak_alias (__tsearch, tsearch)
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#endif
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/* Find datum in search tree.
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   KEY is the key to be located, ROOTP is the address of tree root,
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   COMPAR the ordering function.  */
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void *
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__tfind (key, vrootp, compar)
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     const void *key;
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     void *const *vrootp;
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     __compar_fn_t compar;
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{
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  node *rootp = (node *) vrootp;
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  if (rootp == NULL)
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    return NULL;
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  CHECK_TREE (*rootp);
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  while (*rootp != NULL)
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    {
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      node root = *rootp;
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      int r;
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      r = (*compar) (key, root->key);
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      if (r == 0)
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	return root;
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      rootp = r < 0 ? &root->left : &root->right;
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    }
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  return NULL;
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}
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#ifdef weak_alias
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weak_alias (__tfind, tfind)
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#endif
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/* Delete node with given key.
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   KEY is the key to be deleted, ROOTP is the address of the root of tree,
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   COMPAR the comparison function.  */
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void *
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__tdelete (const void *key, void **vrootp, __compar_fn_t compar)
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{
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  node p, q, r, retval;
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  int cmp;
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  node *rootp = (node *) vrootp;
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  node root, unchained;
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  /* Stack of nodes so we remember the parents without recursion.  It's
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     _very_ unlikely that there are paths longer than 40 nodes.  The tree
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     would need to have around 250.000 nodes.  */
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  int stacksize = 100;
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  int sp = 0;
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  node *nodestack[100];
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  if (rootp == NULL)
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    return NULL;
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  p = *rootp;
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  if (p == NULL)
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    return NULL;
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  CHECK_TREE (p);
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  while ((cmp = (*compar) (key, (*rootp)->key)) != 0)
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    {
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      if (sp == stacksize)
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	abort ();
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      nodestack[sp++] = rootp;
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      p = *rootp;
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      rootp = ((cmp < 0)
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	       ? &(*rootp)->left
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	       : &(*rootp)->right);
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      if (*rootp == NULL)
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	return NULL;
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    }
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  /* This is bogus if the node to be deleted is the root... this routine
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     really should return an integer with 0 for success, -1 for failure
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     and errno = ESRCH or something.  */
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  retval = p;
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  /* We don't unchain the node we want to delete. Instead, we overwrite
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     it with its successor and unchain the successor.  If there is no
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     successor, we really unchain the node to be deleted.  */
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  root = *rootp;
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  r = root->right;
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  q = root->left;
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  if (q == NULL || r == NULL)
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    unchained = root;
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  else
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    {
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      node *parent = rootp, *up = &root->right;
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      for (;;)
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	{
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	  if (sp == stacksize)
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	    abort ();
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	  nodestack[sp++] = parent;
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	  parent = up;
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	  if ((*up)->left == NULL)
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	    break;
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	  up = &(*up)->left;
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	}
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      unchained = *up;
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    }
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  /* We know that either the left or right successor of UNCHAINED is NULL.
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     R becomes the other one, it is chained into the parent of UNCHAINED.  */
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  r = unchained->left;
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  if (r == NULL)
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    r = unchained->right;
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  if (sp == 0)
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    *rootp = r;
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  else
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    {
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      q = *nodestack[sp-1];
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      if (unchained == q->right)
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	q->right = r;
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      else
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	q->left = r;
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    }
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  if (unchained != root)
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    root->key = unchained->key;
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  if (!unchained->red)
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    {
 | 
						|
      /* Now we lost a black edge, which means that the number of black
 | 
						|
	 edges on every path is no longer constant.  We must balance the
 | 
						|
	 tree.  */
 | 
						|
      /* NODESTACK now contains all parents of R.  R is likely to be NULL
 | 
						|
	 in the first iteration.  */
 | 
						|
      /* NULL nodes are considered black throughout - this is necessary for
 | 
						|
	 correctness.  */
 | 
						|
      while (sp > 0 && (r == NULL || !r->red))
 | 
						|
	{
 | 
						|
	  node *pp = nodestack[sp - 1];
 | 
						|
	  p = *pp;
 | 
						|
	  /* Two symmetric cases.  */
 | 
						|
	  if (r == p->left)
 | 
						|
	    {
 | 
						|
	      /* Q is R's brother, P is R's parent.  The subtree with root
 | 
						|
		 R has one black edge less than the subtree with root Q.  */
 | 
						|
	      q = p->right;
 | 
						|
	      if (q->red)
 | 
						|
		{
 | 
						|
		  /* If Q is red, we know that P is black. We rotate P left
 | 
						|
		     so that Q becomes the top node in the tree, with P below
 | 
						|
		     it.  P is colored red, Q is colored black.
 | 
						|
		     This action does not change the black edge count for any
 | 
						|
		     leaf in the tree, but we will be able to recognize one
 | 
						|
		     of the following situations, which all require that Q
 | 
						|
		     is black.  */
 | 
						|
		  q->red = 0;
 | 
						|
		  p->red = 1;
 | 
						|
		  /* Left rotate p.  */
 | 
						|
		  p->right = q->left;
 | 
						|
		  q->left = p;
 | 
						|
		  *pp = q;
 | 
						|
		  /* Make sure pp is right if the case below tries to use
 | 
						|
		     it.  */
 | 
						|
		  nodestack[sp++] = pp = &q->left;
 | 
						|
		  q = p->right;
 | 
						|
		}
 | 
						|
	      /* We know that Q can't be NULL here.  We also know that Q is
 | 
						|
		 black.  */
 | 
						|
	      if ((q->left == NULL || !q->left->red)
 | 
						|
		  && (q->right == NULL || !q->right->red))
 | 
						|
		{
 | 
						|
		  /* Q has two black successors.  We can simply color Q red.
 | 
						|
		     The whole subtree with root P is now missing one black
 | 
						|
		     edge.  Note that this action can temporarily make the
 | 
						|
		     tree invalid (if P is red).  But we will exit the loop
 | 
						|
		     in that case and set P black, which both makes the tree
 | 
						|
		     valid and also makes the black edge count come out
 | 
						|
		     right.  If P is black, we are at least one step closer
 | 
						|
		     to the root and we'll try again the next iteration.  */
 | 
						|
		  q->red = 1;
 | 
						|
		  r = p;
 | 
						|
		}
 | 
						|
	      else
 | 
						|
		{
 | 
						|
		  /* Q is black, one of Q's successors is red.  We can
 | 
						|
		     repair the tree with one operation and will exit the
 | 
						|
		     loop afterwards.  */
 | 
						|
		  if (q->right == NULL || !q->right->red)
 | 
						|
		    {
 | 
						|
		      /* The left one is red.  We perform the same action as
 | 
						|
			 in maybe_split_for_insert where two red edges are
 | 
						|
			 adjacent but point in different directions:
 | 
						|
			 Q's left successor (let's call it Q2) becomes the
 | 
						|
			 top of the subtree we are looking at, its parent (Q)
 | 
						|
			 and grandparent (P) become its successors. The former
 | 
						|
			 successors of Q2 are placed below P and Q.
 | 
						|
			 P becomes black, and Q2 gets the color that P had.
 | 
						|
			 This changes the black edge count only for node R and
 | 
						|
			 its successors.  */
 | 
						|
		      node q2 = q->left;
 | 
						|
		      q2->red = p->red;
 | 
						|
		      p->right = q2->left;
 | 
						|
		      q->left = q2->right;
 | 
						|
		      q2->right = q;
 | 
						|
		      q2->left = p;
 | 
						|
		      *pp = q2;
 | 
						|
		      p->red = 0;
 | 
						|
		    }
 | 
						|
		  else
 | 
						|
		    {
 | 
						|
		      /* It's the right one.  Rotate P left. P becomes black,
 | 
						|
			 and Q gets the color that P had.  Q's right successor
 | 
						|
			 also becomes black.  This changes the black edge
 | 
						|
			 count only for node R and its successors.  */
 | 
						|
		      q->red = p->red;
 | 
						|
		      p->red = 0;
 | 
						|
 | 
						|
		      q->right->red = 0;
 | 
						|
 | 
						|
		      /* left rotate p */
 | 
						|
		      p->right = q->left;
 | 
						|
		      q->left = p;
 | 
						|
		      *pp = q;
 | 
						|
		    }
 | 
						|
 | 
						|
		  /* We're done.  */
 | 
						|
		  sp = 1;
 | 
						|
		  r = NULL;
 | 
						|
		}
 | 
						|
	    }
 | 
						|
	  else
 | 
						|
	    {
 | 
						|
	      /* Comments: see above.  */
 | 
						|
	      q = p->left;
 | 
						|
	      if (q->red)
 | 
						|
		{
 | 
						|
		  q->red = 0;
 | 
						|
		  p->red = 1;
 | 
						|
		  p->left = q->right;
 | 
						|
		  q->right = p;
 | 
						|
		  *pp = q;
 | 
						|
		  nodestack[sp++] = pp = &q->right;
 | 
						|
		  q = p->left;
 | 
						|
		}
 | 
						|
	      if ((q->right == NULL || !q->right->red)
 | 
						|
		       && (q->left == NULL || !q->left->red))
 | 
						|
		{
 | 
						|
		  q->red = 1;
 | 
						|
		  r = p;
 | 
						|
		}
 | 
						|
	      else
 | 
						|
		{
 | 
						|
		  if (q->left == NULL || !q->left->red)
 | 
						|
		    {
 | 
						|
		      node q2 = q->right;
 | 
						|
		      q2->red = p->red;
 | 
						|
		      p->left = q2->right;
 | 
						|
		      q->right = q2->left;
 | 
						|
		      q2->left = q;
 | 
						|
		      q2->right = p;
 | 
						|
		      *pp = q2;
 | 
						|
		      p->red = 0;
 | 
						|
		    }
 | 
						|
		  else
 | 
						|
		    {
 | 
						|
		      q->red = p->red;
 | 
						|
		      p->red = 0;
 | 
						|
		      q->left->red = 0;
 | 
						|
		      p->left = q->right;
 | 
						|
		      q->right = p;
 | 
						|
		      *pp = q;
 | 
						|
		    }
 | 
						|
		  sp = 1;
 | 
						|
		  r = NULL;
 | 
						|
		}
 | 
						|
	    }
 | 
						|
	  --sp;
 | 
						|
	}
 | 
						|
      if (r != NULL)
 | 
						|
	r->red = 0;
 | 
						|
    }
 | 
						|
 | 
						|
  free (unchained);
 | 
						|
  return retval;
 | 
						|
}
 | 
						|
#ifdef weak_alias
 | 
						|
weak_alias (__tdelete, tdelete)
 | 
						|
#endif
 | 
						|
 | 
						|
 | 
						|
/* Walk the nodes of a tree.
 | 
						|
   ROOT is the root of the tree to be walked, ACTION the function to be
 | 
						|
   called at each node.  LEVEL is the level of ROOT in the whole tree.  */
 | 
						|
static void
 | 
						|
internal_function
 | 
						|
trecurse (const void *vroot, __action_fn_t action, int level)
 | 
						|
{
 | 
						|
  const_node root = (const_node) vroot;
 | 
						|
 | 
						|
  if (root->left == NULL && root->right == NULL)
 | 
						|
    (*action) (root, leaf, level);
 | 
						|
  else
 | 
						|
    {
 | 
						|
      (*action) (root, preorder, level);
 | 
						|
      if (root->left != NULL)
 | 
						|
	trecurse (root->left, action, level + 1);
 | 
						|
      (*action) (root, postorder, level);
 | 
						|
      if (root->right != NULL)
 | 
						|
	trecurse (root->right, action, level + 1);
 | 
						|
      (*action) (root, endorder, level);
 | 
						|
    }
 | 
						|
}
 | 
						|
 | 
						|
 | 
						|
/* Walk the nodes of a tree.
 | 
						|
   ROOT is the root of the tree to be walked, ACTION the function to be
 | 
						|
   called at each node.  */
 | 
						|
void
 | 
						|
__twalk (const void *vroot, __action_fn_t action)
 | 
						|
{
 | 
						|
  const_node root = (const_node) vroot;
 | 
						|
 | 
						|
  CHECK_TREE (root);
 | 
						|
 | 
						|
  if (root != NULL && action != NULL)
 | 
						|
    trecurse (root, action, 0);
 | 
						|
}
 | 
						|
#ifdef weak_alias
 | 
						|
weak_alias (__twalk, twalk)
 | 
						|
#endif
 | 
						|
 | 
						|
 | 
						|
#ifdef _LIBC
 | 
						|
 | 
						|
/* The standardized functions miss an important functionality: the
 | 
						|
   tree cannot be removed easily.  We provide a function to do this.  */
 | 
						|
static void
 | 
						|
internal_function
 | 
						|
tdestroy_recurse (node root, __free_fn_t freefct)
 | 
						|
{
 | 
						|
  if (root->left != NULL)
 | 
						|
    tdestroy_recurse (root->left, freefct);
 | 
						|
  if (root->right != NULL)
 | 
						|
    tdestroy_recurse (root->right, freefct);
 | 
						|
  (*freefct) ((void *) root->key);
 | 
						|
  /* Free the node itself.  */
 | 
						|
  free (root);
 | 
						|
}
 | 
						|
 | 
						|
void
 | 
						|
__tdestroy (void *vroot, __free_fn_t freefct)
 | 
						|
{
 | 
						|
  node root = (node) vroot;
 | 
						|
 | 
						|
  CHECK_TREE (root);
 | 
						|
 | 
						|
  if (root != NULL)
 | 
						|
    tdestroy_recurse (root, freefct);
 | 
						|
}
 | 
						|
weak_alias (__tdestroy, tdestroy)
 | 
						|
 | 
						|
#endif /* _LIBC */
 |