98. Validate Binary Search Tree

落日映苍穹つ 2022-03-07 03:42 179阅读 0赞

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   3
    Output: true

Example 2:

5
       / \
      1   4
         / \
        3   6
    Output: false
    Explanation: 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.

难度：medium

题目：给定二叉树，判断其是否为二叉搜索树。

思路：递归，并记录当前结点的取值范围。

Runtime: 0 ms, faster than 100.00% of Java online submissions for Validate Binary Search Tree.  
Memory Usage: 37.6 MB, less than 100.00% of Java online submissions for Validate Binary Search Tree.

/**
     * 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) {
            if (null == root) {
                return true;
            }
            
            return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
        }
    
        public boolean isValidBST(TreeNode root, long minVal, long maxVal) {
            if (null == root) {
                return true;
            }
            
            if (root.val >= maxVal || root.val <= minVal) {
                return false;
            }
            
            return isValidBST(root.left, minVal, root.val) && isValidBST(root.right, root.val, maxVal);
        }
    }