Skip to content

Latest commit

 

History

History
90 lines (66 loc) · 1.94 KB

20210217.md

File metadata and controls

90 lines (66 loc) · 1.94 KB

Algorithm

98. Validate Binary Search Tree

Description

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 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: 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

Solution

/**
 * 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;
        return dfs(root,Long.MIN_VALUE,Long.MAX_VALUE);
    }

    boolean dfs(TreeNode root,long min,long max){
        // 当前节点为空直接返回true
        if(root==null){
            return true;
        }
        // 当前节点比min或者left节点小,或者当前节点比right节点大,返回false
        if(root.val<=min||root.val>=max){
            return false;
        }
        // 递归执行(left,min, 根结点) &&(right,根结点, max)
        return dfs(root.left,min,(int)root.val)&&
            dfs(root.right,(int)root.val,max);
    }
}

Discuss

Review

Tip

Share