# 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   3
Output: 4
Sum of all non-leaf nodes at 0th level is 4.
Sum of all non-leaf nodes at 1st level is 2.
Sum of all non-leaf nodes at 2nd level is 0.
Hence maximum sum is 4

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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 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; ` `} `

## 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
```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : rituraj_jain

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.