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.


                      /   \
                     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.

               /   \
             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++ 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;
    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();
            // 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)
            if (temp->right != NULL)
        // 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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