HackerRank Binary Search Tree : Lowest Common Ancestor solution

In this HackerRank Binary Search Tree: Lowest Common Ancestor Interview preparation kit problem You are given a pointer to the root of the binary search tree and two values v1 and v2. You need to return the lowest common ancestor (LCA) of v1 and v2 in the binary search tree.

Problem solution in Python programming.

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

# 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
'''

def lca(root, v1, v2):
    if (root.info < v1 and root.info > v2) or (root.info > v1 and root.info < v2):
        return root
    elif 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)
    elif root.info == v1 or root.info == v2:
        return root
    
    
  

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)

Problem solution in Java Programming.

import java.util.*;
import java.io.*;

class Node {
    Node left;
    Node right;
    int data;
    
    Node(int data) {
        this.data = data;
        left = null;
        right = null;
    }
}

class Solution {

    /*
    class Node 
        int data;
        Node left;
        Node right;
    */
    

    /*
    class Node 
        int data;
        Node left;
        Node right;
    */
    static Node lca(Node root,int v1,int v2)
{
    //Decide if you have to call rekursively
    //Samller than both
    if(root.data < v1 && root.data < v2){
        return lca(root.right,v1,v2);
    }
    //Bigger than both
    if(root.data > v1 && root.data > v2){
        return lca(root.left,v1,v2);
    }

    //Else solution already found
    return root;
}


    public static Node insert(Node root, int data) {
        if(root == null) {
            return new Node(data);
        } else {
            Node cur;
            if(data <= root.data) {
                cur = insert(root.left, data);
                root.left = cur;
            } else {
                cur = insert(root.right, data);
                root.right = cur;
            }
            return root;
        }
    }

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int t = scan.nextInt();
        Node root = null;
        while(t-- > 0) {
            int data = scan.nextInt();
            root = insert(root, data);
        }
        int v1 = scan.nextInt();
        int v2 = scan.nextInt();
        scan.close();
        Node ans = lca(root,v1,v2);
        System.out.println(ans.data);
    }   
}

Problem solution in C++ programming.

#include <bits/stdc++.h>

using namespace std;

class Node {
    public:
        int data;
        Node *left;
        Node *right;
        Node(int d) {
            data = d;
            left = NULL;
            right = NULL;
        }
};

class Solution {
    public:
        Node* insert(Node* root, int data) {
            if(root == NULL) {
                return new Node(data);
            } else {
                Node* cur;
                if(data <= root->data) {
                    cur = insert(root->left, data);
                    root->left = cur;
                } else {
                    cur = insert(root->right, data);
                    root->right = cur;
               }

               return root;
           }
        }

/*The tree node has data, left child and right child 
class Node {
    int data;
    Node* left;
    Node* right;
};

*/
  /*
Node is defined as 

typedef struct node
{
   int data;
   node * left;
   node * right;
}node;

*/


Node * lca(Node * root, int v1,int v2)
{
    if(root == nullptr) return nullptr;
    int data = root->data;
    if(v1 < data && v2 < data) return lca(root->left, v1, v2);
    if(v1 > data && v2 > data) return lca(root->right, v1, v2);
    return root;
}

}; //End of Solution

int main() {
  
    Solution myTree;
    Node* root = NULL;
    
    int t;
    int data;

    std::cin >> t;

    while(t-- > 0) {
        std::cin >> data;
        root = myTree.insert(root, data);
    }
    
    int v1, v2;
    std::cin >> v1 >> v2;
  
    Node *ans = myTree.lca(root, v1, v2);
    
    std::cout << ans->data;

    return 0;
}

Problem solution in C programming.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

struct node {
    
    int data;
    struct node *left;
    struct node *right;
  
};

struct node* insert( struct node* root, int data ) {
        
    if(root == NULL) {
    
        struct node* node = (struct node*)malloc(sizeof(struct node));

        node->data = data;

        node->left = NULL;
        node->right = NULL;
        return node;
      
    } else {
      
        struct node* cur;
        
        if(data <= root->data) {
            cur = insert(root->left, data);
            root->left = cur;
        } else {
            cur = insert(root->right, data);
            root->right = cur;
        }
    
        return root;
    }
}

/* you only have to complete the function given below.  
node is defined as  

struct node {
    
    int data;
    struct node *left;
    struct node *right;
  
};

*/

struct node *lca( struct node *root, int v1, int v2 ) {

    while(root!= NULL){
        if(v1 > root->data && v2 > root->data){
            root = root->right;
        }else if(v1 < root->data && v2 <  root->data){
            root = root ->left;
        }else{
            break;
        }
    }
    return root;
}


int main() {
  
    struct node* root = NULL;
    
    int t;
    int data;

    scanf("%d", &t);

    while(t-- > 0) {
        scanf("%d", &data);
        root = insert(root, data);
    }
    int v1;
    int v2;
  
    scanf("%d%d", &v1, &v2);
    struct node *ans = lca(root, v1, v2);
    printf("%d", ans->data);
    
    return 0;
}