# Maximum sum of non-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 non-leaf nodes among all level of given binary tree.

**Examples:**

Input:4 / \ 2 -5 / \ -1 3Output:4 Sum of all non-leaf nodes at 0^{th}level is 4. Sum of all non-leaf nodes at 1^{st}level is 2. Sum of all non-leaf nodes at 2^{nd}level is 0. Hence maximum sum is 4Input:1 / \ 2 3 / \ \ 4 5 8 / \ 6 7Output:8

**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 non-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 non-leaf nodes ` `// at any level in tree using level order traversal ` `int` `maxNonLeafNodesSum(` `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 a node from queue ` ` ` `Node* temp = q.front(); ` ` ` `q.pop(); ` ` ` ` ` `// Add non-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 << maxNonLeafNodesSum(root) << endl; ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

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## Python3

# Python3 implementation of the approach

import queue

# A binary tree node has data, pointer to

# left child and a pointer to right child

class Node:

def __init__(self, data):

self.data = data

self.left = None

self.right = None

# Function to return the maximum Sum of

# non-leaf nodes at any level in tree

# using level order traversal

def maxNonLeafNodesSum(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 = queue.Queue()

q.put(root)

while not q.empty():

# Get the size of queue when the level

# order traversal for one level finishes

count = q.qsize()

# Iterate for all the nodes

# in the queue currently

Sum = 0

while count:

# Dequeue a node from queue

temp = q.get()

# Add non-leaf node’s value to current Sum

if temp.left != None or temp.right != None:

Sum += temp.data

# Enqueue left and right

# children of dequeued node

if temp.left != None:

q.put(temp.left)

if temp.right != None:

q.put(temp.right)

count -= 1

# Update the maximum Sum of leaf nodes 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)

print(maxNonLeafNodesSum(root))

# This code is contributed by Rituraj Jain

**Output:**

8

## Recommended Posts:

- Maximum sum of leaf nodes among all levels of the given binary tree
- Print all nodes between two given levels in Binary Tree
- Print Levels of all nodes in a Binary Tree
- Print even positioned nodes of even levels in level order of the given binary tree
- Sum of nodes at maximum depth of a Binary Tree
- Maximum sum of nodes in Binary tree such that no two are adjacent
- Find maximum among all right nodes in Binary Tree
- Sum of nodes at maximum depth of a Binary Tree | Set 2
- Print the nodes at odd levels of a tree
- Maximum sum of nodes in Binary tree such that no two are adjacent | Dynamic Programming
- Sum of nodes at maximum depth of a Binary Tree | Iterative Approach
- Queries to find the maximum Xor value between X and the nodes of a given level of a perfect binary tree
- Maximum length cycle that can be formed by joining two nodes of a binary tree
- Print Binary Tree levels in sorted order | Set 3 (Tree given as array)
- Minimum and maximum node that lies in the path connecting two nodes in a Binary Tree

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