把二元查找树转变成排序的双向链表（）
题目：
输入一棵二元查找树，将该转换成个排序的双向链表。
要求不能创建任何新的结点，只调整指针向。

#include<stdio.h>
#include<stdlib.h>

typedef struct _BitTree {
 struct _BitTree * left;
 struct _BitTree * right;

 int value;
}BitTree;

void AddNode(  BitTree ** root, int value) {
 BitTree *p;
 if( *root == NULL) {
  *root = (BitTree *)malloc(sizeof(BitTree));
  if( *root == NULL)
   return ;
  (*root)->left = NULL;
  (*root)->right = NULL;
  (*root)->value = value;
  return;
 }

 p = *root;
 
 while(1) {

  if(p->value > value){
   if(p->left == NULL) {
    p->left = (BitTree *)malloc(sizeof(BitTree ));
    if(p->left == NULL)
     printf(“malloc left error\n”);
    p->left->value = value;
    p->left->left = NULL;
    p->left->right = NULL;
    return;
   }
   p = p->left;
  }else if(p->value <  value) {
   if(p->right == NULL) {
    p->right = (BitTree *) malloc ( sizeof ( BitTree ));
    if( p->right == NULL)
     printf (“malloc right error\n”);
    p->right->value = value;
    p->right->left = NULL;
    p->right->right = NULL;
    return;
    }
   p = p->right;
  }else {
   printf(“congfu\n”);
   return;
  }
 }
 return;

}

void ConvertTree(BitTree ** head, BitTree ** tail, BitTree* root) {
 BitTree *h = NULL, *t = NULL;

 if(root == NULL)
  return;

 ConvertTree ( &h, &t, root->left);
 if(t != NULL) {
  root->left = t;
  t->right = root;
        *head = h;
 }
 
 h = NULL;
 t = NULL;

 ConvertTree (&h, &t, root->right);
 if(h != NULL) {
  root->right = h;
  h->left = root;
  *tail = t;
 }

 if(h == NULL)
  *head = root;

 if(t == NULL)
  *tail = root;
 
}
  
int main(int argc, char *argv[]) {
 BitTree *tree = NULL, *head = NULL, *tail = NULL;
 AddNode(&tree, 10);
 AddNode(&tree, 6);
 AddNode(&tree, 14);
 AddNode(&tree, 4);
 AddNode(&tree, 3);
 AddNode(&tree, 5);
 AddNode(&tree, 8);
 AddNode(&tree, 7);
 AddNode(&tree, 9);
 AddNode(&tree, 12);
 AddNode(&tree, 11);
 AddNode(&tree, 13);
 AddNode(&tree, 15);
 
 ConvertTree(&head, &tail, tree);

 while(tree->left != NULL)
  tree = tree->left;

 while(tree->right != NULL) {

  printf(“%d “, tree->value);
  tree = tree->right;
 }

 return 0;
}