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)