LeetCode - Validate Binary Search Tree

Problem description

description

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

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

Example 2:

    5
   / \
  1   4
     / \
    3   6

Input: [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.

Analysis

Just use inorder traversal and give a min value and max value to test if it’s BST.

/**
 * 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) {
        return inorder(root, (long)Integer.MIN_VALUE - 1, (long)Integer.MAX_VALUE + 1);
    }
    
    boolean inorder(TreeNode node, long min, long max){
        if (node == null){
            return true;
        }
        
        boolean res = inorder(node.left, min, node.val);
        if (!res){
            return false;
        }
        
        if (node.val <= min || node.val >= max){
            return false;
        }
        
        res = inorder(node.right, node.val, max);
        
        return res;
    }
}

If we don’t want to use long, then we can use Integer instead of int.

class Solution {
  public boolean helper(TreeNode node, Integer lower, Integer upper) {
    if (node == null) return true;

    int val = node.val;
    if (lower != null && val <= lower) return false;
    if (upper != null && val >= upper) return false;

    if (! helper(node.right, val, upper)) return false;
    if (! helper(node.left, lower, val)) return false;
    return true;
  }

  public boolean isValidBST(TreeNode root) {
    return helper(root, null, null);
  }
}

What to improve

• be careful about the Integer overflow problem.