二叉搜索树与双向链表

// 排序二叉树与双向链表.cpp : 定义控制台应用程序的入口点。
// 题目描述：输入一棵二叉搜索树，将该二叉搜索树转换成一个排序的双向链表。要求不能创建任何新的结点，只能调整树中结点指针的指向。
// 思路：通过对一颗排序二叉树的认识可知一颗排序二叉树中序输出就是有序的
//       通过后序遍历让子树分别找到自己的left和right结点，然后再与根节点相结合，
//       结合原则：一个结点的left是左子树的最右孩子，一个结点的right是右子树的最左孩子
//                  通过后序遍历就不必保存中间结果。

#include "stdafx.h"
struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
TreeNode(int x) :
val(x), left(NULL), right(NULL) {
}
};
TreeNode* dfs(TreeNode* pRootOfTree);
TreeNode* findLeft(TreeNode* pRootOfTree); // 声明着大批左节点函数，指向左子树的最右孩子
TreeNode* findRight(TreeNode* pRootOfTree); //声明找到右结点函数，指向右子树的最左孩子
TreeNode* Convert(TreeNode* pRootOfTree)
{
if (pRootOfTree == NULL)
return NULL;
TreeNode* tmp = dfs(pRootOfTree);          //递归的改变树的left和right区域

while (tmp->left != NULL) {
tmp = tmp->left;
}
TreeNode* res = tmp;
/*while (tmp->right != NULL) {
tmp->right->left = tmp;
tmp = tmp->right;
}*/
return res;
}
TreeNode* dfs(TreeNode* pRootOfTree) {
if (pRootOfTree == NULL)
return NULL;

if (pRootOfTree->left != NULL)
dfs(pRootOfTree->left);                //处理左子树
if (pRootOfTree->right != NULL)
dfs(pRootOfTree->right);               //处理右子树

pRootOfTree->left = findLeft(pRootOfTree->left);       //处理自身，找到自己的left，并把找到的right结点的left域指向自己
if (findLeft(pRootOfTree->left) != NULL)
findLeft(pRootOfTree->left)->right = pRootOfTree;

pRootOfTree->right = findRight(pRootOfTree->right);   //处理自身，找到自己的right，并把找到的left结点的right域指向自己
if (findRight(pRootOfTree->right) != NULL)
findRight(pRootOfTree->right)->left = pRootOfTree;
return pRootOfTree;

}
TreeNode* findLeft(TreeNode* pRootOfTree) {              //左子树的最右孩子
TreeNode* tmp = pRootOfTree;
if (tmp == NULL)
return NULL;
while (tmp->right != NULL) {
tmp = tmp->right;
}
return tmp;
}

TreeNode* findRight(TreeNode* pRootOfTree) {  //右子树的最左孩子
TreeNode* tmp = pRootOfTree;
if (tmp == NULL)
return NULL;
while (tmp->left != NULL) {
tmp = tmp->left;
}
return tmp;
}

int main()            //生成测试数据
{
TreeNode *head;
TreeNode *test4 = new TreeNode(4);
TreeNode *test6 = new TreeNode(6);
TreeNode *test8 = new TreeNode(8);
TreeNode *test10 = new TreeNode(10);
TreeNode *test12 = new TreeNode(12);
TreeNode *test14 = new TreeNode(14);
TreeNode *test16 = new TreeNode(16);

test10->left = test6;
test10->right = test14;
test6->left = test4;
test6->right = test8;
test14->left = test12;
test14->right = test16;
TreeNode* res = Convert(test10);
return 0;
}