首页 > 代码库 > 【leetcode刷题笔记】Recover Binary Search Tree

【leetcode刷题笔记】Recover Binary Search Tree

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?


 

题解:需要找到二叉搜索树中乱序的两个节点,并把它们交换回来。

排序二叉树的中序遍历得到的一定是一个递增序列,例如上述所示的BST,中序遍历得到的序列为1,3,5,9,11,15,20。假设错位的是3和11,那么错位后的树如下图所示

   

用遍历firstNode和secondNode表示错位的两个点,那么在中序遍历过程中,第一个错位点后面的点一定比它小(5比11小,11才是错位的点);第二个错位点一定比它前面的点小(3比9小)。采用递归的方法中序遍历BST,如果当前遍历的点root比它前面的点prev的值小,有两种可能,第一种是prev是第一个错位的点,此时firstNode变量为空,则将firstNode赋值为prev(代码中第24行);第二种是root是第二个错位的点,此时firstNode变量部位空,将secondNode变量赋值为root(代码中第22行)。

代码如下:

 1 /** 2  * Definition for binary tree 3  * public class TreeNode { 4  *     int val; 5  *     TreeNode left; 6  *     TreeNode right; 7  *     TreeNode(int x) { val = x; } 8  * } 9  */10 public class Solution {11     private TreeNode firstNode = null;12     private TreeNode secondNode = null;13     private TreeNode prev = null;14     15     private void traverse(TreeNode root){16         if(root == null)17             return;18         19         traverse(root.left);20         21         if(prev != null && root.val < prev.val){22             secondNode = root;23             if(firstNode == null)24                 firstNode = prev;25         }26         prev = root;27         traverse(root.right);28             29     }30     31     public void recoverTree(TreeNode root) {32         traverse(root);33         34         int temp = firstNode.val;35         firstNode.val = secondNode.val;36         secondNode.val = temp;37     }38 }