# Find maximum level sum in Binary Tree

Given a Binary Tree having positive and negative nodes, the task is to find the maximum sum level in it.

**Examples:**

Input : 4 / \ 2 -5 / \ /\ -1 3 -2 6 Output: 6 Explanation : Sum of all nodes of 0'th level is 4 Sum of all nodes of 1'th level is -3 Sum of all nodes of 0'th level is 6 Hence maximum sum is 6 Input : 1 / \ 2 3 / \ \ 4 5 8 / \ 6 7 Output : 17

This problem is a variation of the maximum width problem. The idea is to do a level order traversal of the tree. While doing traversal, process nodes of different levels separately. For every level being processed, compute the sum of nodes in the level and keep track of the maximum sum.

Below is the implementation of the above idea:

## C++

// A queue based C++ program to find maximum sum // of a level in Binary Tree #include <bits/stdc++.h> using namespace std; /* A binary tree node has data, pointer to left child and a pointer to right child */ struct Node { int data; struct Node *left, *right; }; // Function to find the maximum sum of a level in tree // using level order traversal int maxLevelSum(struct Node* root) { // Base case if (root == NULL) return 0; // Initialize result int result = root->data; // Do Level order traversal keeping track of number // of nodes at every level. queue<Node*> q; q.push(root); while (!q.empty()) { // Get the size of queue when the level order // traversal for one level finishes int count = q.size(); // Iterate for all the nodes in the queue currently int sum = 0; while (count--) { // Dequeue an node from queue Node* temp = q.front(); q.pop(); // Add this node's value to current sum. sum = sum + temp->data; // Enqueue left and right children of // dequeued node if (temp->left != NULL) q.push(temp->left); if (temp->right != NULL) q.push(temp->right); } // Update the maximum node count value result = max(sum, result); } return result; } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */ struct Node* newNode(int data) { struct Node* node = new Node; node->data = data; node->left = node->right = NULL; return (node); } // Driver code int main() { struct Node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); root->right->right = newNode(8); root->right->right->left = newNode(6); root->right->right->right = newNode(7); /* Constructed Binary tree is: 1 / \ 2 3 / \ \ 4 5 8 / \ 6 7 */ cout << "Maximum level sum is " << maxLevelSum(root) << endl; return 0; }

## Java

// A queue based Java program to find maximum // sum of a level in Binary Tree import java.util.LinkedList; import java.util.Queue; class GFG{ // A binary tree node has data, pointer // to left child and a pointer to right // child static class Node { int data; Node left, right; public Node(int data) { this.data = data; this.left = this.right = null; } }; // Function to find the maximum // sum of a level in tree // using level order traversal static int maxLevelSum(Node root) { // Base case if (root == null) return 0; // Initialize result int result = root.data; // Do Level order traversal keeping // track of number of nodes at every // level. Queue<Node> q = new LinkedList<>(); q.add(root); while (!q.isEmpty()) { // Get the size of queue when the // level order traversal for one // level finishes int count = q.size(); // Iterate for all the nodes // in the queue currently int sum = 0; while (count-- > 0) { // Dequeue an node from queue Node temp = q.poll(); // Add this node's value // to current sum. sum = sum + temp.data; // Enqueue left and right children // of dequeued node if (temp.left != null) q.add(temp.left); if (temp.right != null) q.add(temp.right); } // Update the maximum node // count value result = Math.max(sum, result); } return result; } // Driver code public static void main(String[] args) { Node root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.left.right = new Node(5); root.right.right = new Node(8); root.right.right.left = new Node(6); root.right.right.right = new Node(7); /* Constructed Binary tree is: 1 / \ 2 3 / \ \ 4 5 8 / \ 6 7 */ System.out.println("Maximum level sum is " + maxLevelSum(root)); } } // This code is contributed by sanjeev2552

## Python3

# A queue based Python3 program to find # maximum sum of a level in Binary Tree from collections import deque # A binary tree node has data, pointer # to left child and a pointer to right # child class Node: def __init__(self, key): self.data = key self.left = None self.right = None # Function to find the maximum sum # of a level in tree # using level order traversal def maxLevelSum(root): # Base case if (root == None): return 0 # Initialize result result = root.data # Do Level order traversal keeping # track of number # of nodes at every level. q = deque() q.append(root) while (len(q) > 0): # Get the size of queue when the # level order traversal for one # level finishes count = len(q) # Iterate for all the nodes in # the queue currently sum = 0 while (count > 0): # Dequeue an node from queue temp = q.popleft() # Add this node's value to current sum. sum = sum + temp.data # Enqueue left and right children of # dequeued node if (temp.left != None): q.append(temp.left) if (temp.right != None): q.append(temp.right) count -= 1 # Update the maximum node count value result = max(sum, result) return result # Driver code if __name__ == '__main__': root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) root.right.right = Node(8) root.right.right.left = Node(6) root.right.right.right = Node(7) # Constructed Binary tree is: # 1 # / \ # 2 3 # / \ \ # 4 5 8 # / \ # 6 7 print("Maximum level sum is", maxLevelSum(root)) # This code is contributed by mohit kumar 29

## C#

// A queue based C# program to find maximum // sum of a level in Binary Tree using System; using System.Collections.Generic; class GFG { // A binary tree node has data, pointer // to left child and a pointer to right // child public class Node { public int data; public Node left, right; public Node(int data) { this.data = data; this.left = this.right = null; } }; // Function to find the maximum // sum of a level in tree // using level order traversal static int maxLevelSum(Node root) { // Base case if (root == null) return 0; // Initialize result int result = root.data; // Do Level order traversal keeping // track of number of nodes at every // level. Queue<Node> q = new Queue<Node>(); q.Enqueue(root); while (q.Count != 0) { // Get the size of queue when the // level order traversal for one // level finishes int count = q.Count; // Iterate for all the nodes // in the queue currently int sum = 0; while (count --> 0) { // Dequeue an node from queue Node temp = q.Dequeue(); // Add this node's value // to current sum. sum = sum + temp.data; // Enqueue left and right children // of dequeued node if (temp.left != null) q.Enqueue(temp.left); if (temp.right != null) q.Enqueue(temp.right); } // Update the maximum node // count value result = Math.Max(sum, result); } return result; } // Driver code public static void Main(String[] args) { Node root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.left.right = new Node(5); root.right.right = new Node(8); root.right.right.left = new Node(6); root.right.right.right = new Node(7); /* Constructed Binary tree is: 1 / \ 2 3 / \ \ 4 5 8 / \ 6 7 */ Console.WriteLine("Maximum level sum is " + maxLevelSum(root)); } } // This code is contributed by gauravrajput1

## Javascript

<script> // A queue based Javascript program to find maximum // sum of a level in Binary Tree // A binary tree node has data, pointer // to left child and a pointer to right // child class Node { constructor(data) { this.data = data; this.left = this.right = null; } } // Function to find the maximum // sum of a level in tree // using level order traversal function maxLevelSum(root) { // Base case if (root == null) return 0; // Initialize result let result = root.data; // Do Level order traversal keeping // track of number of nodes at every // level. let q = []; q.push(root); while (q.length!=0) { // Get the size of queue when the // level order traversal for one // level finishes let count = q.length; // Iterate for all the nodes // in the queue currently let sum = 0; while (count-- > 0) { // Dequeue an node from queue let temp = q.shift(); // Add this node's value // to current sum. sum = sum + temp.data; // Enqueue left and right children // of dequeued node if (temp.left != null) q.push(temp.left); if (temp.right != null) q.push(temp.right); } // Update the maximum node // count value result = Math.max(sum, result); } return result; } // Driver code let root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.left.right = new Node(5); root.right.right = new Node(8); root.right.right.left = new Node(6); root.right.right.right = new Node(7); /* Constructed Binary tree is: 1 / \ 2 3 / \ \ 4 5 8 / \ 6 7 */ document.write("Maximum level sum is " + maxLevelSum(root)); // This code is contributed by unknown2108 </script>

**Output**

Maximum level sum is 17

__Complexity Analysis:__

**Time Complexity**: O(N) where N is the total number of nodes in the tree.

In level order traversal, every node of the tree is processed once, and hence the complexity due to the level order traversal is O(N) if there are total N nodes in the tree. Also, while processing every node, we are maintaining the sum at each level, however, this does not affect the overall time complexity. Therefore, the time complexity is O(N).

**Auxiliary Space:** O(w) where w is the maximum width of the tree.

In level order traversal, a queue is maintained whose maximum size at any moment can go up to the maximum width of the binary tree.

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