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:

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    2
   / \
  1   3
Binary tree [2,1,3], return true.

Example 2:

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    1
   / \
  2   3
Binary tree [1,2,3], return false.

O(n) time O(1) space recursive solution

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isValidBST(TreeNode root) {
        return isValid(root, null, null);
    }
    
    private boolean isValid(TreeNode root, Integer lower, Integer upper) {
        if (root != null) {
            return (lower == null || root.val > lower) && (upper == null || root.val < upper) && isValid(root.left, lower, root.val) && isValid(root.right, root.val, upper);
        }
        return true;
    }
}

Iterative Inorder traversal solution

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public boolean isValidBST (TreeNode root){
		   Stack<TreeNode> stack = new Stack<TreeNode> ();
		   TreeNode cur = root ;
		   TreeNode pre = null ;		   
		   while (!stack.isEmpty() || cur != null) {			   
			   if (cur != null) {
				   stack.push(cur);
				   cur = cur.left ;
			   } else {				   
				   TreeNode p = stack.pop() ;
				   if (pre != null && p.val <= pre.val) {					   
					   return false ;
				   }				   
				   pre = p ;					   
				   cur = p.right ;
			   }
		   }
		   return true ; 
	   }