[LeetCode] 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.

 Thought process:
An inorder traversal of a binary search tree will result in a sorted list. The idea is to perform an inorder traversal on the tree, and use an instance variable to update the previously visited node. If the previous node's value is larger than current node's, the tree is not a BST.

 Solution 1 (recursive):
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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private TreeNode previous = null;
    
    public boolean isValidBST(TreeNode root) {
        if (root == null) {
            return true;
        }
        if (!isValidBST(root.left)) {
            return false;
        }
        if (previous != null && root.val <= previous.val) {
            return false;
        }
        previous = root;
        return isValidBST(root.right);
    }
}

Solution 2 (iterative):

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isValidBST(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        TreeNode previous = null;
        
        while (!stack.isEmpty() || root != null) {
            while (root != null) {
                stack.push(root);
                root = root.left;
            }
            
            root = stack.pop();
            if (previous != null && previous.val >= root.val) {
                return false;
            }
            previous = root;
            root = root.right;
        }
        return true;
    }
}

Time complexity: O(n).
Location: Philadelphia, PA, USA

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