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.

Source: leetcode | Difficulty: MEDIUM_HARD
Tags: Tree algorithms

Graph traversal

Range queries

Starter code: # Definition for a binary tree node.
class TreeNode:
    def __init__(self, val=0, left=None, right=None):
    self.val = val
    self.left = left
    self.right = right
class Solution:
    def isValidBST(self, root: TreeNode) -> bool:
        

Solutions:
class Solution:

	def isValidBST(self, root):
		if not root:
			return True
		return self.dfs(root)[2]

	def dfs(self, node):
		if node is None:
			return [None, None, True]
		L = self.dfs(node.left)
		R = self.dfs(node.right)
		if not L[2] or not R[2] or (L[1] is not None and L[1] >= node.val) or (R[0] is not None and R[0] <= node.val):
			return [None, None, False]
		if L[0] is not None:
			min = L[0]
		else:
			min = node.val
		if R[1] is not None:
			max = R[1]
		else:
			max = node.val
		return [min, max, True]


class Solution(object):

	def isValidBST(self, root):
		res = []
		return self.inorder(root, res)

	def inorder(self, root, res):
		if not root:
			return True
		if self.inorder(root.left, res) == False:
			return False
		if len(res) != 0:
			last = res.pop()
			if last >= root.val:
				return False
			res.append(last)
		res.append(root.val)
		if self.inorder(root.right, res) == False:
			return False
		return True