把二元查找树转变成排序的双向链表

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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;
}

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