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:
Input:
    2
   / \
  1   3
Output: true
Example 2:
    5
   / \
  1   4
     / \
    3   6
Output: false
Explanation: The input is: [5,1,4,null,null,3,6]. The root node's value
             is 5 but its right child's value is 4.
# Definition for a binary tree node.
 # class TreeNode:
 #     def __init__(self, x):
 #         self.val = x
 #         self.left = None
 #         self.right = None
 
 class Solution:
     def isValidBST(self, root):
         """
         :type root: TreeNode
         :rtype: bool
         """
         stack = []
         pre = None
         while stack or root:
             while root:
                 stack.append(root)
                 root = root.left
             root = stack.pop()
             if pre and root.val <= pre.val: return False
             pre = root
             root = root.right
         return True
     
# Definition for a binary tree node.
 # class TreeNode:
 #     def __init__(self, x):
 #         self.val = x
 #         self.left = None
 #         self.right = None
 
 class Solution:
     def isValidBST(self, root):
         """
         :type root: TreeNode
         :rtype: bool
         """
         stack = []
         pre = None
         while stack or root:
             while root:
                 stack.append(root)
                 root = root.left
             root = stack.pop()
             if pre and root.val <= pre.val: return False
             pre = root
             root = root.right
         return True
     