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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;
    }
};