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.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    // BST inorder traversal.
    public boolean isValidBST(TreeNode root) {
        if (root == null) return true;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        TreeNode pre = null;
        while (!stack.isEmpty() || cur != null) {
            if (cur != null) {
                stack.push(cur);
                cur = cur.left;
            } else {
                TreeNode node = stack.pop();
                if (pre != null && node.val <= pre.val) {
                    return false;
                } 
                pre = node;
                cur = node.right;
            }
        }
        return true;
    }
    
    /**
    // Recursion
    public boolean isValidBST(TreeNode root) {
        long upper = Integer.MAX_VALUE;
        long lower = Integer.MIN_VALUE;
        return isValid(root, upper + 1, lower - 1);
    }
    private boolean isValid(TreeNode root, long upper, long lower) {
        if (root == null) return true;
        return root.val < upper && root.val > lower &&
               isValid(root.left, root.val, lower) &&
               isValid(root.right, upper, root.val);
    }
    */
}
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