【leetcode刷题笔记】Validate Binary Search Tree

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

The left subtree of a node contains only nodes with keys less than the node's key.
The right subtree of a node contains only nodes with keys greater than the node's key.
Both the left and right subtrees must also be binary search trees.

题解：BST valid 的充分必要条件是它的中序遍历是一个有序序列。

递归实现树的中序遍历，用私有变量lastVal记录上一个遍历的节点的值。在一次递归，首先递归判断左子树是否是BST，并且更新lastVal，然后将root的值跟lastVal比较，看root的值是否大于lastVal；然后递归判断右子树是否是BST。

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 int lastVal = Integer.MIN_VALUE;
12     public boolean isValidBST(TreeNode root) {
13         if(root == null)
14             return true;
15         
16         if(!isValidBST(root.left))
17             return false;
18         
19         if(root.val <= lastVal)
20             return false;
21         lastVal = root.val;
22         if(!isValidBST(root.right))
23             return false;
24         return true;
25     }
26 }

题目的关键点是lastVal更新的时机和与root比较的时机。