Binary search tree or not

https://github.com/Premiumlab/Python-for-Algorithms--Data-Structures--and-Interviews/blob/master/Mock%20Interviews/Ride%20Share%20Start-Up%20Company/Ride%20Share%20Company%20-%20Interview%20Questions%20-%20SOLUTIONS/On-Site%20Question%203%20-%20SOLUTION.ipynb

 

 

 

On-Site Question 3 - SOLUTION

Problem

Given a binary tree, check whether it’s a binary search tree or not.

 

Requirements

Use paper/pencil, do not code this in an IDE until you've done it manually

Do not use built-in Python libraries to do this, but do mention them if you know about them

 

Solution

The first solution that comes to mind is, at every node check whether its value is larger than or equal to its left child and smaller than or equal to its right child (assuming equals can appear at either left or right). However, this approach is erroneous because it doesn’t check whether a node violates any condition with its grandparent or any of its ancestors.

So, we should keep track of the minimum and maximum values a node can take. And at each node we will check whether its value is between the min and max values it’s allowed to take. The root can take any value between negative infinity and positive infinity. At any node, its left child should be smaller than or equal than its own value, and similarly the right child should be larger than or equal to. So during recursion, we send the current value as the new max to our left child and send the min as it is without changing. And to the right child, we send the current value as the new min and send the max without changing.

 

 

class Node: 
    def __init__(self, val=None): 
        self.left, self.right, self.val = None, None, val   
        
INFINITY = float("infinity") 
NEG_INFINITY = float("-infinity")  

def isBST(tree, minVal=NEG_INFINITY, maxVal=INFINITY): 
    if tree is None:
        return True   
    if not minVal <= tree.val <= maxVal: 
        return False   
    
    return isBST(tree.left, minVal, tree.val) and isBST(tree.right, tree.val, maxVal)

 

 

 

There’s an equally good alternative solution. If a tree is a binary search tree, then traversing the tree inorder should lead to sorted order of the values in the tree. So, we can perform an inorder traversal and check whether the node values are sorted or not.

 

 

def isBST2(tree, lastNode=[NEG_INFINITY]): 
    
    if tree is None: 
        return True   
    
    if not isBST2(tree.left, lastNode):
        return False   
    
    if tree.val < lastNode[0]: 
        return False   
    
    lastNode[0]=tree.val   
    
    return isBST2(tree.right, lastNode)

 

 

This is a common interview problem, its relatively simple, but not trivial and shows that someone has a knowledge of binary search trees and tree traversals.

Good Job!