# Maximum sum of leaf nodes among all levels of the given binary tree

• Difficulty Level : Easy
• Last Updated : 25 Jun, 2021

Given a Binary Tree having positive and negative nodes, the task is to find the maximum sum of leaf nodes among all level of the given binary tree.
Examples:

```Input:
4
/   \
2    -5
/ \
-1   3
Output: 2
Sum of all leaves at 0th level is 0.
Sum of all leaves at 1st level is -5.
Sum of all leaves at 2nd level is 2.
Hence maximum sum is 2.

Input:
1
/   \
2      3
/  \      \
4    5      8
/   \
6     7
Output: 13```

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Approach: The idea to solve the above problem is to do level order traversal of tree. While doing traversal, process nodes of different level separately. For every level being processed, compute the sum of leaf nodes in the level and keep track of the maximum sum.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``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 return the maximum sum of leaf nodes``// at any level in tree using level order traversal``int` `maxLeafNodesSum(``struct` `Node* root)``{` `    ``// Base case``    ``if` `(root == NULL)``        ``return` `0;` `    ``// Initialize result``    ``int` `result = 0;` `    ``// Do Level order traversal keeping track``    ``// of the number of nodes at every level``    ``queue 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 leaf node's value to current sum``            ``if` `(temp->left == NULL && temp->right == NULL)` `                ``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 sum of leaf nodes 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);``    ``cout << maxLeafNodesSum(root) << endl;` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;``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;``};` `// Function to return the maximum sum``// of leaf nodes at any level in tree``// using level order traversal``static` `int` `maxLeafNodesSum(Node root)``{` `    ``// Base case``    ``if` `(root == ``null``)``        ``return` `0``;` `    ``// Initialize result``    ``int` `result = ``0``;` `    ``// Do Level order traversal keeping track``    ``// of the number of nodes at every level``    ``Queue 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.peek();``            ``q.remove();` `            ``// Add leaf node's value to current sum``            ``if` `(temp.left == ``null` `&&``                ``temp.right == ``null``)` `                ``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 sum of leaf nodes value``        ``result = Math.max(sum, result);``    ``}` `    ``return` `result;``}` `// Helper function that allocates a new node with the``// given data and null left and right pointers``static` `Node newNode(``int` `data)``{``    ``Node node = ``new` `Node();``    ``node.data = data;``    ``node.left = node.right = ``null``;``    ``return` `(node);``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``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``);``    ``System.out.println(maxLeafNodesSum(root));``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 implementation of the approach` `# A binary tree node has data,``# pointer to left child and``# a pointer to right child``# Helper function that allocates``# a new node with the given data``# and None left and right pointers``class` `newNode:``    ``def` `__init__(``self``, data):``        ``self``.data ``=` `data``        ``self``.left ``=` `None``        ``self``.right ``=` `None` `# Function to return the maximum sum``# of leaf nodes at any level in tree``# using level order traversal``def` `maxLeafNodesSum(root):``    ` `    ``# Base case``    ``if` `(root ``=``=` `None``):``        ``return` `0``        ` `    ``# Initialize result``    ``result ``=` `0``    ` `    ``# Do Level order traversal keeping track``    ``# of the number of nodes at every level``    ``q ``=` `[]``    ``q.append(root)``    ``while``(``len``(q)):``        ` `        ``# 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):``            ` `            ``# Dequeue an node from queue``            ``temp ``=` `q[``0``]``            ``q.pop(``0``)` `            ``# Add leaf node's value to current sum``            ``if` `(temp.left ``=``=` `None` `and``                ``temp.right ``=``=` `None``):``                ``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 sum``        ``# of leaf nodes value``        ``result ``=` `max``(``sum``, result)``    ` `    ``return` `result``        ` `# Driver code``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``)``print``(maxLeafNodesSum(root))` `# This code is contributed by SHUBHAMSINGH10`

## C#

 `// C# implementation of the approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `// A binary tree node has data,``// pointer to left child and``// a pointer to right child``class` `Node``{``    ``public` `int` `data;``    ``public` `Node left, right;``};` `// Function to return the maximum sum``// of leaf nodes at any level in tree``// using level order traversal``static` `int` `maxLeafNodesSum(Node root)``{` `    ``// Base case``    ``if` `(root == ``null``)``        ``return` `0;` `    ``// Initialize result``    ``int` `result = 0;` `    ``// Do Level order traversal keeping track``    ``// of the number of nodes at every level``    ``Queue q = ``new` `Queue();``    ``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.Peek();``            ``q.Dequeue();` `            ``// Add leaf node's value to current sum``            ``if` `(temp.left == ``null` `&&``                ``temp.right == ``null``)` `                ``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 sum of leaf nodes value``        ``result = Math.Max(sum, result);``    ``}` `    ``return` `result;``}` `// Helper function that allocates a new node with the``// given data and null left and right pointers``static` `Node newNode(``int` `data)``{``    ``Node node = ``new` `Node();``    ``node.data = data;``    ``node.left = node.right = ``null``;``    ``return` `(node);``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``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);``    ``Console.WriteLine(maxLeafNodesSum(root));``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``
Output:
`13`

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