把二元查找树转变成排序的双向链表

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;

}