Sum of all nodes at Kth level in a Binary Tree

Given a binary tree with N nodes and an integer K, the task is to find the sum of all the nodes present at the K^th level.

Examples:

Input:

K = 1 
Output: 70 

Input:

K = 2 
Output: 120 

Approach:

• Traverse the Binary Tree using Level Order Traversal and queue
• During traversal, pop each element out of the queue and push it’s child (if available) in the queue.
• Keep the track of the current level of the Binary tree.
• To track the current level, declare a variable level and increase it whenever a child is traversed from the parent.
• When the current level of the tree i.e. the variable level meets the required K^th level, pop the elements from the queue and calculate their sum.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach

#include <bits/stdc++.h>
using namespace std;

// Binary tree node consists of data, a
// pointer to the left child and a
// pointer to the right child
struct node {
    int data;
    struct node* left;
    struct node* right;
};

// Function to create new Binary Tree node
struct node* newNode(int data)
{
    struct node* temp = new struct node;
    temp->data = data;
    temp->left = nullptr;
    temp->right = nullptr;
    return temp;
};

// Function to return the sum of all
// the nodes at Kth level using
// level order traversal
int sumOfNodesAtNthLevel(struct node* root,
                         int k)
{

    // If the current node is NULL
    if (root == nullptr)
        return 0;

    // Create Queue
    queue<struct node*> que;

    // Enqueue the root node
    que.push(root);

    // Level is used to track
    // the current level
    int level = 0;

    // To store the sum of nodes
    // at the Kth level
    int sum = 0;

    // flag is used to break out of
    // the loop after the sum of all
    // the nodes at Nth level is found
    int flag = 0;

    // Iterate the queue till its not empty
    while (!que.empty()) {

        // Calculate the number of nodes
        // in the current level
        int size = que.size();

        // Process each node of the current
        // level and enqueue their left
        // and right child to the queue
        while (size--) {
            struct node* ptr = que.front();
            que.pop();

            // If the current level matches the
            // required level then calculate the
            // sum of all the nodes at that level
            if (level == k) {

                // Flag initialized to 1
                // indicates that sum of the
                // required level is calculated
                flag = 1;

                // Calculating the sum of the nodes
                sum += ptr->data;
            }
            else {

                // Traverse to the left child
                if (ptr->left)
                    que.push(ptr->left);

                // Traverse to the right child
                if (ptr->right)
                    que.push(ptr->right);
            }
        }

        // Increment the variable level
        // by 1 for each level
        level++;

        // Break out from the loop after the sum
        // of nodes at K level is found
        if (flag == 1)
            break;
    }
    return sum;
}

// Driver code
int main()
{
    struct node* root = new struct node;

    // Tree Construction
    root = newNode(50);
    root->left = newNode(30);
    root->right = newNode(70);
    root->left->left = newNode(20);
    root->left->right = newNode(40);
    root->right->left = newNode(60);
    int level = 2;
    int result = sumOfNodesAtNthLevel(root, level);

    // Printing the result
    cout << result;

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

// Binary tree node consists of data, a
// pointer to the left child and a
// pointer to the right child
static class node
{
    int data;
    node left;
    node right;
};

// Function to create new Binary Tree node
static node newNode(int data)
{
    node temp = new node();
    temp.data = data;
    temp.left = null;
    temp.right = null;
    return temp;
};

// Function to return the sum of all
// the nodes at Kth level using
// level order traversal
static int sumOfNodesAtNthLevel(node root,
                                int k)
{

    // If the current node is null
    if (root == null)
        return 0;

    // Create Queue
    Queue<node> que = new LinkedList<>();

    // Enqueue the root node
    que.add(root);

    // Level is used to track
    // the current level
    int level = 0;

    // To store the sum of nodes
    // at the Kth level
    int sum = 0;

    // flag is used to break out of
    // the loop after the sum of all
    // the nodes at Nth level is found
    int flag = 0;

    // Iterate the queue till its not empty
    while (!que.isEmpty())
    {

        // Calculate the number of nodes
        // in the current level
        int size = que.size();

        // Process each node of the current
        // level and enqueue their left
        // and right child to the queue
        while (size-- >0)
        {
            node ptr = que.peek();
            que.remove();

            // If the current level matches the
            // required level then calculate the
            // sum of all the nodes at that level
            if (level == k)
            {

                // Flag initialized to 1
                // indicates that sum of the
                // required level is calculated
                flag = 1;

                // Calculating the sum of the nodes
                sum += ptr.data;
            }
            else {

                // Traverse to the left child
                if (ptr.left != null)
                    que.add(ptr.left);

                // Traverse to the right child
                if (ptr.right != null)
                    que.add(ptr.right);
            }
        }

        // Increment the variable level
        // by 1 for each level
        level++;

        // Break out from the loop after the sum
        // of nodes at K level is found
        if (flag == 1)
            break;
    }
    return sum;
}

// Driver code
public static void main(String[] args)
{
    node root = new node();

    // Tree Construction
    root = newNode(50);
    root.left = newNode(30);
    root.right = newNode(70);
    root.left.left = newNode(20);
    root.left.right = newNode(40);
    root.right.left = newNode(60);
    int level = 2;
    int result = sumOfNodesAtNthLevel(root, level);

    // Printing the result
    System.out.print(result);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach

# Binary tree node consists of data, a
# pointer to the left child and a
# pointer to the right child
class newNode :
    def __init__(self, data) :
        self.data = data;
        self.left = None;
        self.right = None;

# Function to return the sum of all
# the nodes at Kth level using
# level order traversal
def sumOfNodesAtNthLevel(root, k) :

    # If the current node is NULL
    if (root == None) :
        return 0;

    # Create Queue
    que = [];

    # Enqueue the root node
    que.append(root);

    # Level is used to track
    # the current level
    level = 0;

    # To store the sum of nodes
    # at the Kth level
    sum = 0;

    # flag is used to break out of
    # the loop after the sum of all
    # the nodes at Nth level is found
    flag = 0;

    # Iterate the queue till its not empty
    while (len(que) != 0) :

        # Calculate the number of nodes
        # in the current level
        size = len(que);

        # Process each node of the current
        # level and enqueue their left
        # and right child to the queue
        while (size != 0) :

            size -= 1;
            ptr = que[0];
            que.pop(0);

            # If the current level matches the
            # required level then calculate the
            # sum of all the nodes at that level
            if (level == k) :

                # Flag initialized to 1
                # indicates that sum of the
                # required level is calculated
                flag = 1;

                # Calculating the sum of the nodes
                sum += ptr.data;

            else :
                # Traverse to the left child
                if (ptr.left) :
                    que.append(ptr.left);

                # Traverse to the right child
                if (ptr.right) :
                    que.append(ptr.right);

        # Increment the variable level
        # by 1 for each level
        level += 1;

        # Break out from the loop after the sum
        # of nodes at K level is found
        if (flag == 1) :
            break;

    return sum;

# Driver code
if __name__ == "__main__" :

    # Tree Construction
    root = newNode(50);
    root.left = newNode(30);
    root.right = newNode(70);
    root.left.left = newNode(20);
    root.left.right = newNode(40);
    root.right.left = newNode(60);
    level = 2;
    result = sumOfNodesAtNthLevel(root, level);

    # Printing the result
    print(result);

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;

class GFG
{

// Binary tree node consists of data, a
// pointer to the left child and a
// pointer to the right child
class node
{
    public int data;
    public node left;
    public node right;
};

// Function to create new Binary Tree node
static node newNode(int data)
{
    node temp = new node();
    temp.data = data;
    temp.left = null;
    temp.right = null;
    return temp;
}

// Function to return the sum of all
// the nodes at Kth level using
// level order traversal
static int sumOfNodesAtNthLevel(node root,
                                int k)
{

    // If the current node is null
    if (root == null)
        return 0;

    // Create Queue
    List<node> que = new List<node>();

    // Enqueue the root node
    que.Add(root);

    // Level is used to track
    // the current level
    int level = 0;

    // To store the sum of nodes
    // at the Kth level
    int sum = 0;

    // flag is used to break out of
    // the loop after the sum of all
    // the nodes at Nth level is found
    int flag = 0;

    // Iterate the queue till its not empty
    while (que.Count != 0)
    {

        // Calculate the number of nodes
        // in the current level
        int size = que.Count;

        // Process each node of the current
        // level and enqueue their left
        // and right child to the queue
        while (size-- >0)
        {
            node ptr = que[0];
            que.RemoveAt(0);

            // If the current level matches the
            // required level then calculate the
            // sum of all the nodes at that level
            if (level == k)
            {

                // Flag initialized to 1
                // indicates that sum of the
                // required level is calculated
                flag = 1;

                // Calculating the sum of the nodes
                sum += ptr.data;
            }
            else
            {

                // Traverse to the left child
                if (ptr.left != null)
                    que.Add(ptr.left);

                // Traverse to the right child
                if (ptr.right != null)
                    que.Add(ptr.right);
            }
        }

        // Increment the variable level
        // by 1 for each level
        level++;

        // Break out from the loop after the sum
        // of nodes at K level is found
        if (flag == 1)
            break;
    }
    return sum;
}

// Driver code
public static void Main(String[] args)
{
    node root = new node();

    // Tree Construction
    root = newNode(50);
    root.left = newNode(30);
    root.right = newNode(70);
    root.left.left = newNode(20);
    root.left.right = newNode(40);
    root.right.left = newNode(60);
    int level = 2;
    int result = sumOfNodesAtNthLevel(root, level);

    // Printing the result
    Console.Write(result);
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// JavaScript implementation of the approach

// Binary tree node consists of data, a
// pointer to the left child and a
// pointer to the right child
class node
{
    constructor()
    {
        this.data = 0;
        this.left = null;
        this.right = null;
    }
};

// Function to create new Binary Tree node
function newNode(data)
{
    var temp = new node();
    temp.data = data;
    temp.left = null;
    temp.right = null;
    return temp;
}

// Function to return the sum of all
// the nodes at Kth level using
// level order traversal
function sumOfNodesAtNthLevel(root, k)
{

    // If the current node is null
    if (root == null)
        return 0;

    // Create Queue
    var que = [];

    // Enqueue the root node
    que.push(root);

    // Level is used to track
    // the current level
    var level = 0;

    // To store the sum of nodes
    // at the Kth level
    var sum = 0;

    // flag is used to break out of
    // the loop after the sum of all
    // the nodes at Nth level is found
    var flag = 0;

    // Iterate the queue till its not empty
    while (que.length != 0)
    {

        // Calculate the number of nodes
        // in the current level
        var size = que.length;

        // Process each node of the current
        // level and enqueue their left
        // and right child to the queue
        while (size-- >0)
        {
            var ptr = que[0];
            que.shift();

            // If the current level matches the
            // required level then calculate the
            // sum of all the nodes at that level
            if (level == k)
            {

                // Flag initialized to 1
                // indicates that sum of the
                // required level is calculated
                flag = 1;

                // Calculating the sum of the nodes
                sum += ptr.data;
            }
            else
            {

                // Traverse to the left child
                if (ptr.left != null)
                    que.push(ptr.left);

                // Traverse to the right child
                if (ptr.right != null)
                    que.push(ptr.right);
            }
        }

        // Increment the variable level
        // by 1 for each level
        level++;

        // Break out from the loop after the sum
        // of nodes at K level is found
        if (flag == 1)
            break;
    }
    return sum;
}

// Driver code
var root = new node();
// Tree Construction
root = newNode(50);
root.left = newNode(30);
root.right = newNode(70);
root.left.left = newNode(20);
root.left.right = newNode(40);
root.right.left = newNode(60);
var level = 2;
var result = sumOfNodesAtNthLevel(root, level);
// Printing the result
document.write(result);


</script>

Output:

120

Time Complexity: O(N)

Auxiliary Space: O(N) because using auxiliary space for queue que

Approach 2: Recursive DFS

In this approach, we can traverse the tree in a Depth First Search (DFS) manner, and maintain a level count. Whenever we reach a node at the kth level, we add its value to a sum. We continue this process for all nodes in the tree and return the final sum.

    Initialize a variable sum to 0.
   Define a recursive function that takes in a root node, a current level, and a target level:
   a. If the current node is null, return 0.
   b. If the current level equals the target level, add the value of the current node to the sum.
   c. Recursively call the function on the left subtree with an incremented level.
   d. Recursively call the function on the right subtree with an incremented level.
   e. Return the sum.
   Call the recursive function with the root node, the starting level of 1, and the target level.

Here’s the C++ code:

C++

#include <iostream>
using namespace std;

// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

// Function to calculate the sum of nodes at kth level in a binary tree
int sumAtKthLevel(TreeNode* root, int k, int level = 0) {
    if (root == NULL) {
        return 0;
    }
    if (level == k) {
        return root->val;
    }
    return sumAtKthLevel(root->left, k, level + 1) + sumAtKthLevel(root->right, k, level + 1);
}

// Driver code
int main() {
    // Example binary tree
    TreeNode* root = new TreeNode(50);
    root->left = new TreeNode(30);
    root->right = new TreeNode(70);
    root->left->left = new TreeNode(20);
    root->left->right = new TreeNode(40);
    root->right->left = new TreeNode(60);


    int k = 2;
    int sum = sumAtKthLevel(root, k);

    cout << "The sum of nodes at level " << k << " is " << sum << endl;

    return 0;
}

Output

The sum of nodes at level 2 is 120

Time complexity: O(n), where n is the number of nodes in the tree. We visit every node once.

Space complexity: O(h), where h is the height of the tree. The maximum amount of space used in the call stack is equal to the maximum depth of the recursion, which is the height of the tree.