【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.

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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。