Skip to content
Related Articles

Related Articles

Improve Article

Sum of cousins of a given node in a Binary Tree

  • Difficulty Level : Medium
  • Last Updated : 24 Jun, 2021

Given a binary tree and data value of a node. The task is to find the sum of cousin nodes of given node. If given node has no cousins then return -1. 
Note: It is given that all nodes have distinct values and the given node exists in the tree. 
Examples: 
 

Input: 
                1
              /  \
             3    7
           /  \  / \
          6   5  4  13
             /  / \
            10 17 15
         key = 13
Output: 11
Cousin nodes are 5 and 6 which gives sum 11. 

Input:
                1
              /  \
             3    7
           /  \  / \
          6   5  4  13
             /  / \
            10 17 15
           key = 7
Output: -1
No cousin nodes of node having value 7.

 

Approach: The approach is to do a level order traversal of the tree. While performing level order traversal, find the sum of child nodes of next level. Add a child node’s value to the sum and check if either of the children nodes is the target node or not. If yes, then do not add the value of either child to the sum. After traversing current level if the target node is present in next level, then end the level order traversal and sum found is the sum of cousin nodes. 
Below is the implementation of the above approach: 
 

C++




// CPP program to find sum of cousins
// of given node in binary tree.
#include <bits/stdc++.h>
using namespace std;
 
// A Binary Tree Node
struct Node {
    int data;
    struct Node *left, *right;
};
 
// A utility function to create a new
// Binary Tree Node
struct Node* newNode(int item)
{
    struct Node* temp = (struct Node*)malloc(sizeof(struct Node));
    temp->data = item;
    temp->left = temp->right = NULL;
    return temp;
}
 
// Function to find sum of cousins of
// a given node.
int findCousinSum(Node* root, int key)
{
    if (root == NULL)
        return -1;
 
    // Root node has no cousins so return -1.
    if (root->data == key) {
        return -1;
    }
 
    // To store sum of cousins.
    int currSum = 0;
 
    // To store size of current level.
    int size;
 
    // To perform level order traversal.
    queue<Node*> q;
    q.push(root);
 
    // To represent that target node is
    // found.
    bool found = false;
 
    while (!q.empty()) {
 
        // If target node is present at
        // current level, then return
        // sum of cousins stored in currSum.
        if (found == true) {
            return currSum;
        }
 
        // Find size of current level and
        // traverse entire level.
        size = q.size();
        currSum = 0;
 
        while (size) {
            root = q.front();
            q.pop();
 
            // Check if either of the existing
            // children of given node is target
            // node or not. If yes then set
            // found equal to true.
            if ((root->left && root->left->data == key)
                || (root->right && root->right->data == key)) {
                found = true;
            }
 
            // If target node is not children of
            // current node, then its childeren can be cousin
            // of target node, so add their value to sum.
            else {
                if (root->left) {
                    currSum += root->left->data;
                    q.push(root->left);
                }
 
                if (root->right) {
                    currSum += root->right->data;
                    q.push(root->right);
                }
            }
 
            size--;
        }
    }
 
    return -1;
}
 
// Driver Code
int main()
{
    /*
                1
              /  \
             3    7
           /  \  / \
          6   5  4  13
             /  / \
            10 17 15
    */
 
    struct Node* root = newNode(1);
    root->left = newNode(3);
    root->right = newNode(7);
    root->left->left = newNode(6);
    root->left->right = newNode(5);
    root->left->right->left = newNode(10);
    root->right->left = newNode(4);
    root->right->right = newNode(13);
    root->right->left->left = newNode(17);
    root->right->left->right = newNode(15);
 
    cout << findCousinSum(root, 13) << "\n";
 
    cout << findCousinSum(root, 7) << "\n";
    return 0;
}

Java




// Java program to find sum of cousins
// of given node in binary tree.
import java.util.*;
class Sol
{
     
// A Binary Tree Node
static class Node
{
    int data;
    Node left, right;
};
 
// A utility function to create a new
// Binary Tree Node
static Node newNode(int item)
{
    Node temp = new Node();
    temp.data = item;
    temp.left = temp.right = null;
    return temp;
}
 
// Function to find sum of cousins of
// a given node.
static int findCousinSum(Node root, int key)
{
    if (root == null)
        return -1;
 
    // Root node has no cousins so return -1.
    if (root.data == key)
    {
        return -1;
    }
 
    // To store sum of cousins.
    int currSum = 0;
 
    // To store size of current level.
    int size;
 
    // To perform level order traversal.
    Queue<Node> q=new LinkedList<Node>();
    q.add(root);
 
    // To represent that target node is
    // found.
    boolean found = false;
 
    while (q.size() > 0)
    {
 
        // If target node is present at
        // current level, then return
        // sum of cousins stored in currSum.
        if (found == true)
        {
            return currSum;
        }
 
        // Find size of current level and
        // traverse entire level.
        size = q.size();
        currSum = 0;
 
        while (size > 0)
        {
            root = q.peek();
            q.remove();
 
            // Check if either of the existing
            // children of given node is target
            // node or not. If yes then set
            // found equal to true.
            if ((root.left!=null && root.left.data == key)
                || (root.right!=null && root.right.data == key))
            {
                found = true;
            }
 
            // If target node is not children of
            // current node, then its childeren can be cousin
            // of target node, so add their value to sum.
            else
            {
                if (root.left != null)
                {
                    currSum += root.left.data;
                    q.add(root.left);
                }
 
                if (root.right != null)
                {
                    currSum += root.right.data;
                    q.add(root.right);
                }
            }
 
            size--;
        }
    }
 
    return -1;
}
 
// Driver Code
public static void main(String args[])
{
    /*
                1
            / \
            3 7
        / \ / \
        6 5 4 13
            / / \
            10 17 15
    */
 
    Node root = newNode(1);
    root.left = newNode(3);
    root.right = newNode(7);
    root.left.left = newNode(6);
    root.left.right = newNode(5);
    root.left.right.left = newNode(10);
    root.right.left = newNode(4);
    root.right.right = newNode(13);
    root.right.left.left = newNode(17);
    root.right.left.right = newNode(15);
 
    System.out.print( findCousinSum(root, 13) + "\n");
 
    System.out.print( findCousinSum(root, 7) + "\n");
}
}
 
// This code is contributed by Arnab Kundu

Python3




""" Python3 program to find sum of cousins
of given node in binary tree """
 
# A Binary Tree Node
# Utility function to create a new tree node
class newNode:
 
    # Constructor to create a newNode
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
 
# Function to find sum of cousins of
# a given node.
def findCousinSum( root, key):
 
    if (root == None):
        return -1
 
    # Root node has no cousins so return -1.
    if (root.data == key):
        return -1
     
 
    # To store sum of cousins.
    currSum = 0
 
    # To store size of current level.
    size = 0
 
    # To perform level order traversal.
    q = []
    q.append(root)
 
    # To represent that target node is
    # found.
    found = False
 
    while (len(q)):
 
        # If target node is present at
        # current level, then return
        # sum of cousins stored in currSum.
        if (found == True):
            return currSum
         
        # Find size of current level and
        # traverse entire level.
        size = len(q)
        currSum = 0
 
        while (size):
            root = q[0]
            q.pop(0)
 
            # Check if either of the existing
            # children of given node is target
            # node or not. If yes then set
            # found equal to true.
            if ((root.left and root.left.data == key) or
                (root.right and root.right.data == key)) :
                found = True
             
            # If target node is not children of current
            # node, then its childeren can be cousin
            # of target node, so add their value to sum.
            else:
                if (root.left):
                    currSum += root.left.data
                    q.append(root.left)
                 
                if (root.right) :
                    currSum += root.right.data
                    q.append(root.right)
 
            size -= 1
    return -1
                         
# Driver Code
if __name__ == '__main__':
 
    """
                1
            / \
            3 7
        / \ / \
        6 5 4 13
            / / \
            10 17 15
    """
    root = newNode(1)
    root.left = newNode(3)
    root.right = newNode(7)
    root.left.left = newNode(6)
    root.left.right = newNode(5)
    root.left.right.left = newNode(10)
    root.right.left = newNode(4)
    root.right.right = newNode(13)
    root.right.left.left = newNode(17)
    root.right.left.right = newNode(15)
 
    print(findCousinSum(root, 13))
 
    print(findCousinSum(root, 7))
 
# This code is contributed by
# SHUBHAMSINGH10

C#




// C# program to find sum of cousins
// of given node in binary tree.
using System;
using System.Collections.Generic;
 
class Sol
{
     
// A Binary Tree Node
public class Node
{
    public int data;
    public Node left, right;
};
 
// A utility function to create a new
// Binary Tree Node
static Node newNode(int item)
{
    Node temp = new Node();
    temp.data = item;
    temp.left = temp.right = null;
    return temp;
}
 
// Function to find sum of cousins of
// a given node.
static int findCousinSum(Node root, int key)
{
    if (root == null)
        return -1;
 
    // Root node has no cousins so return -1.
    if (root.data == key)
    {
        return -1;
    }
 
    // To store sum of cousins.
    int currSum = 0;
 
    // To store size of current level.
    int size;
 
    // To perform level order traversal.
    Queue<Node> q = new Queue<Node>();
    q.Enqueue(root);
 
    // To represent that target node is
    // found.
    bool found = false;
 
    while (q.Count > 0)
    {
 
        // If target node is present at
        // current level, then return
        // sum of cousins stored in currSum.
        if (found == true)
        {
            return currSum;
        }
 
        // Find size of current level and
        // traverse entire level.
        size = q.Count;
        currSum = 0;
 
        while (size > 0)
        {
            root = q.Peek();
            q.Dequeue();
 
            // Check if either of the existing
            // children of given node is target
            // node or not. If yes then set
            // found equal to true.
            if ((root.left != null && root.left.data == key)
                || (root.right != null && root.right.data == key))
            {
                found = true;
            }
 
            // If target node is not children of
            // current node, then its childeren can be cousin
            // of target node, so add their value to sum.
            else
            {
                if (root.left != null)
                {
                    currSum += root.left.data;
                    q.Enqueue(root.left);
                }
 
                if (root.right != null)
                {
                    currSum += root.right.data;
                    q.Enqueue(root.right);
                }
            }
 
            size--;
        }
    }
 
    return -1;
}
 
// Driver Code
public static void Main(String []args)
{
    /*
                1
            / \
            3 7
        / \ / \
        6 5 4 13
            / / \
            10 17 15
    */
 
    Node root = newNode(1);
    root.left = newNode(3);
    root.right = newNode(7);
    root.left.left = newNode(6);
    root.left.right = newNode(5);
    root.left.right.left = newNode(10);
    root.right.left = newNode(4);
    root.right.right = newNode(13);
    root.right.left.left = newNode(17);
    root.right.left.right = newNode(15);
 
    Console.Write( findCousinSum(root, 13) + "\n");
 
    Console.Write( findCousinSum(root, 7) + "\n");
}
}
 
// This code is contributed by Princi Singh

Javascript




<script>
 
// JavaScript program to find sum of cousins
// of given node in binary tree.
 
// A Binary Tree Node
class Node
{
  constructor()
  {
    this.data = 0;
    this.left = null;
    this.right = null;
  }
};
 
// A utility function to create a new
// Binary Tree Node
function newNode(item)
{
    var temp = new Node();
    temp.data = item;
    temp.left = temp.right = null;
    return temp;
}
 
// Function to find sum of cousins of
// a given node.
function findCousinSum(root, key)
{
    if (root == null)
        return -1;
 
    // Root node has no cousins so return -1.
    if (root.data == key)
    {
        return -1;
    }
 
    // To store sum of cousins.
    var currSum = 0;
 
    // To store size of current level.
    var size;
 
    // To perform level order traversal.
    var q = [];
    q.push(root);
 
    // To represent that target node is
    // found.
    var found = false;
 
    while (q.length > 0)
    {
 
        // If target node is present at
        // current level, then return
        // sum of cousins stored in currSum.
        if (found == true)
        {
            return currSum;
        }
 
        // Find size of current level and
        // traverse entire level.
        size = q.length;
        currSum = 0;
 
        while (size > 0)
        {
            root = q[0];
            q.shift();
 
            // Check if either of the existing
            // children of given node is target
            // node or not. If yes then set
            // found equal to true.
            if ((root.left != null && root.left.data == key)
                || (root.right != null && root.right.data == key))
            {
                found = true;
            }
 
            // If target node is not children of
            // current node, then its childeren can be cousin
            // of target node, so add their value to sum.
            else
            {
                if (root.left != null)
                {
                    currSum += root.left.data;
                    q.push(root.left);
                }
 
                if (root.right != null)
                {
                    currSum += root.right.data;
                    q.push(root.right);
                }
            }
 
            size--;
        }
    }
 
    return -1;
}
 
// Driver Code
/*
            1
        / \
        3 7
    / \ / \
    6 5 4 13
        / / \
        10 17 15
*/
var root = newNode(1);
root.left = newNode(3);
root.right = newNode(7);
root.left.left = newNode(6);
root.left.right = newNode(5);
root.left.right.left = newNode(10);
root.right.left = newNode(4);
root.right.right = newNode(13);
root.right.left.left = newNode(17);
root.right.left.right = newNode(15);
document.write( findCousinSum(root, 13) + "<br>");
document.write( findCousinSum(root, 7) + "<br>");
 
</script>
Output: 
11
-1

 

Time Complexity: O(N) 
Auxiliary Space: O(N)
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :