Letcode 98 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.

Example 1:

    2   / \  1   3

Binary tree [2,1,3], return true.

Example 2:

    1   / \  2   3

Binary tree [1,2,3], return false.判断是否是BST，有一个地方要注意的是

左子树所有的value都比root小，反之亦然

我觉得自己写的真是太啰嗦了

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

看看漂亮的写法

private boolean help(TreeNode p, Integer low, Integer high) {    if (p == null) return true;    return (low == null || p.val > low) && (high == null || p.val < high) && help(p.left, low, p.val) && help(p.right, p.val, high);}public boolean isValidBST(TreeNode root) {    return help(root, null, null);}