Why are few Testcases of finding Valid BST are failing here in python?


So, I am trying to find ,Given the root of a binary tree, determine if it is a valid binary search tree (BST).

A valid 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.

So, my isssues is few test cases are passing in it but few are not for example following test is failing enter image description here

Following is my python code, can you optimise it better?

class TreeNode(object):
     def __init__(self, val=0, left=None, right=None):
         self.val = val
         self.left = left
         self.right = right
class Solution(object):
    def isValidBST(self, root):
        if not root:
            return []
        if self.isValidBST(root.left)< [root.val] < self.isValidBST(root.right):
            return True
            return False

So, I have written the code but it’s failing test cases with input like root = [2,1,3], output false but explected output is true. It’s passing for test cases like root = [5,1,4,null,null,3,6]

Asked By: vilnius19



The main problem with this implementation is that isValidBST is supposed to return a boolean, so the following pieces of code don’t make sense:

  • return []: that should be return True
  • self.isValidBST(root.left)< [root.val]: this would be comparing a boolean with a list, which is not supported (nor does it have a meaning). Even if you expected the function to return a list here, that list would be empty (as you only return [] as a list — see previous point).

A way to make it work is to pass extra (optional) arguments to isValidBST that provide a "window" (range) for the values of the given tree: all its nodes should have values within that window or it is not a BST. Initially that window can be set with default values to -Infinity to Infinity.

Here is a spoiler in case you cannot make it work:

class Solution(object):
def isValidBST(self, root, low=float('-inf'), high=float('inf')):
return (not root or
low < root.val < high and self.isValidBST(root.left, low, root.val)
and self.isValidBST(root.right, root.val, high))

Alternatively you could make a helper function that does not return a boolean, but returns the window (range) that a given subtree actually spans (so its minimum and maximum values). If either span retrieved from the two subtrees violates the current root node’s value, it is not a BST. But it is important to see that this function must be a different function from the main function, since the main function must return a boolean.

Yet another alternative is to perform an inorder traversal. Each visited value must be greater than the previously visited value.

Answered By: trincot