Leetcode--Recover Binary Search Tree

Problem Description:

Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

Note:
A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1  / \ 2   3    /   4    \     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

分析：题目意思是在一个儿叉搜索树中有两个数字的位置颠倒了，要求找出来并恢复正常的顺序，很容易想到的一种做法是利用中序遍历存储下来所有节点的指针，然后遍历这个中序序列找到两个颠倒的节点，然后进行交换，下面的这个思路是在网上看到的，直接中序遍历的时候每次和前面的一个节点进行比较，找到两个颠倒的节点p1，p2，然后进行交换，不需要先将中序遍历保存下来然后遍历，具体代码如下：

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode *pre,*p1,*p2;        void helper(TreeNode *root)    {        if(root==NULL)            return;        helper(root->left);        if(pre&&pre->val>root->val)//找到第一对逆序的两个数        {            if(p1==NULL)//先找到p1            {                p1=pre;                p2=root;            }                    else//如果找到第二对逆序的两个数则是p2            {                p2=root;            }        }        pre=root;//先将pre初始化为root        helper(root->right);    }    void recoverTree(TreeNode *root) {        if(root==NULL)            return;        pre=p1=p2=NULL;        helper(root);        int temp;        temp=p1->val;        p1->val=p2->val;        p2->val=temp;        return;    }};