Leetcode--Recover Binary Search Tree

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：題目意思是在一個兒叉搜索樹中有兩個數字的位置顛倒了，要求找出來並恢復正常的順序，很容易想到的一種做法是利用中序遍歷存儲下來所有節點的指針，然後遍歷這個中序序列找到兩個顛倒的節點，然後進行交換，下面的這個思路是在網上看到的，直接中序遍歷的時候每次和前面的一個節點進行比較，找到兩個顛倒的節點p1，p2，然後進行交換，不需要先將中序遍歷保存下來然後遍歷，具體代碼如下：

/**
 * 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)//找到第一對逆序的兩個數
        {
            if(p1==NULL)//先找到p1
            {
                p1=pre;
                p2=root;
            }
        
            else//如果找到第二對逆序的兩個數則是p2
            {
                p2=root;
            }
        }
        pre=root;//先將pre初始化為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;
    }
};