lcaBST.py

class Node:
    def __init__(self, info): 
        self.info = info  
        self.left = None  
        self.right = None 
        self.level = None 

    def __str__(self):
        return str(self.info) 

class BinarySearchTree:
    def __init__(self): 
        self.root = None

    def create(self, val):  
        if self.root == None:
            self.root = Node(val)
        else:
            current = self.root
         
            while True:
                if val < current.info:
                    if current.left:
                        current = current.left
                    else:
                        current.left = Node(val)
                        break
                elif val > current.info:
                    if current.right:
                        current = current.right
                    else:
                        current.right = Node(val)
                        break
                else:
                    break

# https://www.hackerrank.com/challenges/binary-search-tree-lowest-common-ancestor/problem?h_l=interview&playlist_slugs%5B%5D=interview-preparation-kit&playlist_slugs%5B%5D=trees&h_r=next-challenge&h_v=zen
# Enter your code here. Read input from STDIN. Print output to STDOUT
'''
class Node:
      def __init__(self,info): 
          self.info = info  
          self.left = None  
          self.right = None 
           

       // this is a node of the tree , which contains info as data, left , right
'''

# INSIGHT COMES FROM THE FACT THAT LCA MUST BE BETWEEN V1 & V2 
def lca(root, v1, v2):
    #Enter your code here
    if root.info < v1 and root.info < v2:
        return lca(root.right, v1, v2)
    
    elif root.info > v1 and root.info > v2:
        return lca(root.left, v1, v2)

    return root
                
if __name__ == '__main__':
    tree = BinarySearchTree()
    t = int(input())

    arr = list(map(int, input().split()))

    for i in range(t):
        tree.create(arr[i])

    v = list(map(int, input().split()))

    ans = lca(tree.root, v[0], v[1])
    print (ans.info)