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.

Basic knowledge

  1. Binary Tree – Inorder Traversal

Solution

We can solve this problem by doing inorder traversal in iterative way.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public bool IsValidBST(TreeNode root) {
        if(root == null)
            return true;
        TreeNode pre = null;
        Stack<TreeNode> s = new Stack<TreeNode>();
        while(root != null || s.Count > 0)
        {
            while(root != null)
            {
                s.Push(root);
                root = root.left;
            }
            root = s.Pop();
            if(pre != null && pre.val >= root.val)
                return false;
            pre = root;
            root = root.right;
        }
        return true;
    }
}

Time Complexity: O(n)

Space Complexity: O(n)