二刷98. Validate Binary Search Tree

Medium
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:

    2
   / \
  1   3

Binary tree [2,1,3], return true.

Example 2:

    1
   / \
  2   3

Binary tree [1,2,3], return false.

二刷还是看答案做出来的，这道题有几个坑：

• root的val可能超出Integer的范围，比如有一个test case给的root.val = 2147483647
• 要注意BST的定义里，每个节点左边的子数val必须小于该节点，右边的子树必须大于该节点。不只是直接的left, right, 而是左边所有的、右边所有的，这个概念可能会有模糊理解

/**
 * 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) {
        if (root == null){
            return true;
        }
        return helper(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }
    
    private boolean helper(TreeNode root, long low, long high){
        if (root.val <= low || root.val >= high){
            return false;
        }
        if (root.left != null && !helper(root.left, low, root.val)){
            return false;
        }
        if (root.right != null && !helper(root.right, root.val, high)){
            return false;
        }
        return true;
    }
}

这道题还可以用inorder traversal做：