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Check if two nodes in a Binary Tree are siblings
  • Difficulty Level : Easy
  • Last Updated : 03 Jun, 2021

Given a binary tree and two nodes, the task is to check if the nodes are siblings of each other or not.

Two nodes are said to be siblings if they are present at the same level, and their parents are same.

Examples: 

Input : 
       1
      /  \
     2    3
    / \  / \
   4   5 6  7
First node is 4 and Second node is 6.
Output : No, they are not siblings.

Input :
         1
        /  \
       5    6
      /    /  \
     7     3   4
First node is 3 and Second node is 4
Output : Yes

Approach: On observing carefully, it can be concluded that any node in a binary tree can have maximum of two child nodes. So, since the parent of two siblings must be same, so the idea is to simply traverse the tree and for every node check if the two given nodes are its children. If it is true for any node in the tree then print YES otherwise print NO.

Below is the implementation of the above approach:  



C++




// C++ program to check if two nodes are
// siblings
 
#include <bits/stdc++.h>
using namespace std;
 
// Binary Tree Node
struct Node {
    int data;
    Node *left, *right;
};
 
// Utility function to create a new node
struct Node* newNode(int data)
{
    struct Node* node = new Node;
    node->data = data;
    node->left = node->right = NULL;
 
    return (node);
}
 
// Function to find out if two nodes are siblings
bool CheckIfNodesAreSiblings(Node* root, int data_one,
                             int data_two)
{
    if (!root)
        return false;
 
    // Compare the two given nodes with
    // the childrens of current node
    if (root->left && root->right) {
        int left = root->left->data;
        int right = root->right->data;
 
        if (left == data_one && right == data_two)
            return true;
        else if (left == data_two && right == data_one)
            return true;
    }
 
    // Check for left subtree
    if (root->left)
        if(CheckIfNodesAreSiblings(root->left, data_one,
                                data_two))
          return true;
 
    // Check for right subtree
    if (root->right)
        if(CheckIfNodesAreSiblings(root->right, data_one,
                                data_two))
          return true;
}
 
// Driver code
int main()
{
 
    struct Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->right->left = newNode(5);
    root->right->right = newNode(6);
    root->left->left->right = newNode(7);
 
    int data_one = 5;
    int data_two = 6;
 
    if (CheckIfNodesAreSiblings(root, data_one, data_two))
        cout << "YES";
    else
        cout << "NO";
 
    return 0;
}

Java




// Java program to check if two nodes
// are siblings
import java.util.*;
 
class GFG{
 
// Binary Tree Node
static class Node
{
    int data;
    Node left, right;
};
 
// Utility function to create a
// new node
static Node newNode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = node.right = null;
 
    return (node);
}
static Node root = null;
 
// Function to find out if two nodes are siblings
static boolean CheckIfNodesAreSiblings(Node root,
                                       int data_one,
                                       int data_two)
{
    if (root == null)
        return false;
 
    // Compare the two given nodes with
    // the childrens of current node
    if (root.left != null && root.right != null)
    {
        int left = root.left.data;
        int right = root.right.data;
 
        if (left == data_one &&
           right == data_two)
            return true;
        else if (left == data_two &&
                right == data_one)
            return true;
    }
 
    // Check for left subtree
    if (root.left != null)
        CheckIfNodesAreSiblings(root.left,
                                data_one,
                                data_two);
 
    // Check for right subtree
    if (root.right != null)
        CheckIfNodesAreSiblings(root.right,
                                data_one,
                                data_two);
    return true;
}
 
// Driver code
public static void main(String[] args)
{
    root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.right.left = newNode(5);
    root.right.right = newNode(6);
    root.left.left.right = newNode(7);
 
    int data_one = 5;
    int data_two = 6;
 
    if (CheckIfNodesAreSiblings(root,
                                data_one,
                                data_two))
        System.out.print("YES");
    else
        System.out.print("NO");
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 program to check if two
# nodes are siblings
 
# Binary Tree Node
class Node:
     
    def __init__(self, key):
         
        self.data = key
        self.left = None
        self.right = None
 
# Function to find out if two nodes are siblings
def CheckIfNodesAreSiblings(root, data_one,
                                  data_two):
     
    if (root == None):
        return False
 
    # Compare the two given nodes with
    # the childrens of current node
    ans = False
     
    if (root.left != None and root.right != None):
        left = root.left.data
        right = root.right.data
         
        if (left == data_one and right == data_two):
            return True
        elif (left == data_two and right == data_one):
            return True
 
    # Check for left subtree
    if (root.left != None):
        ans |= CheckIfNodesAreSiblings(root.left,
                                       data_one,
                                       data_two)
                                        
    # Check for right subtree
    if (root.right != None):
        ans |= CheckIfNodesAreSiblings(root.right,
                                       data_one,
                                       data_two)
         
    return ans   
 
# Driver code
if __name__ == '__main__':
     
    root = Node(1)
    root.left = Node(2)
    root.right = Node(3)
    root.left.left = Node(4)
    root.right.left = Node(5)
    root.right.right = Node(6)
    root.left.left.right = Node(7)
 
    data_one = 5
    data_two = 6
 
    if (CheckIfNodesAreSiblings(root,
                                data_one,
                                data_two)):
        print("YES")
    else:
        print("NO")
 
# This code is contributed by mohit kumar 29

C#




// C# program to check if two nodes
// are siblings
using System;
 
// Binary Tree Node
public class Node
{
    public int data;
    public Node left, right;
     
    // Utility function to create a
    // new node
    public Node(int item)
    {
        data = item;
        left = right = null;
    }
}
 
class GFG{
     
static Node root = null;
 
// Function to find out if two
// nodes are siblings
static bool CheckIfNodesAreSiblings(Node root,
                                    int data_one,
                                    int data_two)
{
    if (root == null)
    {
        return false;
    }
     
    // Compare the two given nodes with
    // the childrens of current node
    if (root.left != null && root.right != null)
    {
        int left = root.left.data;
        int right = root.right.data;
         
        if (left == data_one && right == data_two)
        {
            return true;
        }
        else if (left == data_two && right == data_one)
        {
            return true;
        }
    }
     
    // Check for left subtree
    if (root.left != null)
    {
        CheckIfNodesAreSiblings(root.left, data_one,
                                data_two);
    }
     
    // Check for right subtree
    if (root.right != null)
    {
        CheckIfNodesAreSiblings(root.right, data_one,
                                data_two);
    }
    return true;
}
 
// Driver code
static public void Main()
{
    GFG.root = new Node(1);
    GFG.root.left = new Node(2);
    GFG.root.right = new Node(3);
    GFG.root.left.left = new Node(4);
    GFG.root.right.left = new Node(5);
    GFG.root.right.right = new Node(6);
    GFG.root.left.left.right = new Node(7);
     
    int data_one = 5;
    int data_two = 6;
     
    if (CheckIfNodesAreSiblings(root, data_one,
                                data_two))
    {
        Console.WriteLine("YES");
    }
    else
    {
        Console.WriteLine("NO");
    }
}
}
 
// This code is contributed by avanitrachhadiya2155

Javascript




<script>
 
// Javascript program to check if two nodes
// are siblings
 
// Binary Tree Node
class Node
{
    constructor(data)
    {
        this.data = data;
        this.left = this.right = null;
    }
}
 
let root = null;
 
// Function to find out if two nodes are siblings
function CheckIfNodesAreSiblings(
    root, data_one, data_two)
{
    if (root == null)
        return false;
  
    // Compare the two given nodes with
    // the childrens of current node
    if (root.left != null && root.right != null)
    {
        let left = root.left.data;
        let right = root.right.data;
  
        if (left == data_one &&
           right == data_two)
            return true;
             
        else if (left == data_two &&
                right == data_one)
            return true;
    }
  
    // Check for left subtree
    if (root.left != null)
        CheckIfNodesAreSiblings(root.left,
                                data_one,
                                data_two);
  
    // Check for right subtree
    if (root.right != null)
        CheckIfNodesAreSiblings(root.right,
                                data_one,
                                data_two);
    return true;
}
 
// Driver code
root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.right.left = new Node(5);
root.right.right = new Node(6);
root.left.left.right = new Node(7);
 
let data_one = 5;
let data_two = 6;
 
if (CheckIfNodesAreSiblings(root,
                            data_one,
                            data_two))
    document.write("YES");
else
    document.write("NO");
 
// This code is contributed by unknown2108
 
</script>
Output: 
YES

 

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