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

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

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++

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// 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);
}
  
// Drive 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;
}

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Java

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// 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

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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

Output:

13


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