98. Validate Binary Search Tree

难度: Medium

题目描述

Given the root of a binary tree, determine if it is a valid binary search tree (BST).

A valid BST is defined as follows:

• The left subtree of a node contains only nodes with keys strictly less than the node's key.
• The right subtree of a node contains only nodes with keys strictly greater than the node's key.
• Both the left and right subtrees must also be binary search trees.

 

Example 1:

Input: root = [2,1,3]
Output: true

Example 2:

Input: root = [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.

 

Constraints:

• The number of nodes in the tree is in the range [1, 104].
• -231 <= Node.val <= 231 - 1

解题思路

代码实现

解决方案

java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {

    public boolean isValidBST(TreeNode root) {

        if (root == null) {
            return true;
        }
        boolean left = travalTree(root.left, root, true);
        boolean right = travalTree(root.right, root, false);
        return left && right && isValidBST(root.left) && isValidBST(root.right);

    }

    public boolean travalTree(TreeNode node, TreeNode rootVal, boolean less) {
        if (node == null) {
            return true;
        }
        if (less) {
            if (node.val >= rootVal.val) {
                return false;
            }
        }else{
             if (node.val <= rootVal.val) {
                return false;
            }
        }
        return travalTree(node.left, rootVal, less) && travalTree(node.right, rootVal, less);

    }
}