Validate Binary Search Tree in Python

Suppose we have a binary tree, determine to check whether it is a valid binary search tree (BST) or not. Assume a BST is defined as follows –

  • The left subtree of a node holds only nodes with keys smaller than the node's key.
  • The right subtree of a node holds only nodes with keys larger than the node's key.
  • Both the left and right subtrees must also be binary search trees.

So if the tree is like –

The output will be true.

To solve this, we will follow these steps –

  • Create one recursive function called solve(), this will take root, min and max, the method will be like
  • if root is null, then return true
  • if value of root <= min or value of root >= max, then return false
  • return the (solve(left of root, min, root value) AND solve(right of root, root value, max))
  • call the solve() method initially, by passing root, and – inf as min and inf as max.


Let us see the following implementation to get a better understanding −

 Live Demo

class TreeNode:
   def __init__(self, data, left = None, right = None): = data
      self.left = left
      self.right = right
def insert(temp,data):
   que = []
   while (len(que)):
      temp = que[0]
      if (not temp.left):
         if data is not None:
            temp.left = TreeNode(data)
            temp.left = TreeNode(0)
      if (not temp.right):
         if data is not None:
            temp.right = TreeNode(data)
            temp.right = TreeNode(0)
def make_tree(elements):
   Tree = TreeNode(elements[0])
   for element in elements[1:]:
      insert(Tree, element)
   return Tree
class Solution(object):
   def isValidBST(self, root):
      return self.solve(root,-1000000000000000000000,1000000000000000000000)
   def solve(self,root,min_val,max_val):
      if root == None or == 0:
         return True
      if ( <= min_val or >=max_val):
         return False
      return self.solve(root.left,min_val, and self.solve(root.right,,max_val)
ob1 = Solution()
tree = make_tree([3,1,4,None,2,None,5])
tree = make_tree([5,1,4,None,None,3,6])