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

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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 0^th level is 0.
Sum of all leaves at 1^st level is -5.
Sum of all leaves at 2^nd level is 2.
Hence maximum sum is 2.

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

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 <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 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<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 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<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.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<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.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

<script>
 
    // JavaScript implementation of the approach
     
    // A binary tree node has data, 
    // pointer to left child and 
    // a pointer to right child 
    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }
     
    // Function to return the maximum sum 
    // of leaf nodes at any level in tree 
    // using level order traversal 
    function maxLeafNodesSum(root) 
    { 
 
        // Base case 
        if (root == null) 
            return 0; 
 
        // Initialize result 
        let result = 0; 
 
        // Do Level order traversal keeping track 
        // of the 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[0]; 
                q.shift(); 
 
                // 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 = Math.max(sum, result); 
        } 
 
        return result; 
    } 
 
    // Helper function that allocates a new node with the 
    // given data and null left and right pointers 
    function newNode(data) 
    { 
        let node = new Node(data); 
        return (node); 
    } 
     
    let 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); 
    document.write(maxLeafNodesSum(root));
     
</script>

Output

Time complexity: O(N) where N is the number of node in the given binary tree.
Auxiliary Space: O(N) due to queue data structure.

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