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.

Example 1:

Input:    2   / \  1   3Output: true

Example 2:

5   / \  1   4     / \    3   6Output: falseExplanation: The input is: [5,1,4,null,null,3,6]. The root node's value             is 5 but its right child's value is 4.

 

recursion

给每个节点定义一个区间，如果节点的值落在区间外，返回false，否则继续递归左右节点，区间上下限相应调整

注意！！min, max如果用int会溢出，要用long

time: O(n), space: O(n)

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */class Solution {    public boolean isValidBST(TreeNode root) {        return isValid(root, Long.MIN_VALUE, Long.MAX_VALUE);    }        public boolean isValid(TreeNode node, long min, long max) {        if(node == null) {            return true;        }                if(node.val <= min || node.val >= max) {            return false;        }        return isValid(node.left, min, node.val) && isValid(node.right, node.val, max);    }}