# 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 levels of the 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```

Approach: The idea to solve the above problem is to do the level order traversal of the tree. While doing traversal, process nodes of different levels 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;` `}`

## Java

 `// Java implementation of ` `// the above approach` `import` `java.util.LinkedList;` `import` `java.util.Queue;` `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;` `  ``public` `Node(``int` `data) ` `  ``{` `    ``this``.data = data;` `    ``this``.left = ``this``.right = ``null``;` `  ``}` `};`   `// Function to return the maximum ` `// sum of non-leaf nodes  at any ` `// level in tree using level ` `// order traversal` `static` `int` `maxNonLeafNodesSum(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` `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 a node ` `      ``// from queue` `      ``Node temp = q.poll();`   `      ``// 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.add(temp.left);` `      ``if` `(temp.right != ``null``)` `        ``q.add(temp.right);` `    ``}`   `    ``// Update the maximum sum ` `    ``// of leaf nodes value` `    ``result = max(sum, result);` `  ``}` `  ``return` `result;` `}`   `static` `int` `max(``int` `sum, ` `               ``int` `result) ` `{` `  ``if` `(sum > result)` `    ``return` `sum;` `  ``return` `result;` `}`   `// Driver code` `public` `static` `void` `main(String[] args) ` `{` `  ``Node root = ``new` `Node(``1``);` `  ``root.left = ``new` `Node(``2``);` `  ``root.right = ``new` `Node(``3``);` `  ``root.left.left = ``new` `Node(``4``);` `  ``root.left.right = ``new` `Node(``5``);` `  ``root.right.right = ``new` `Node(``8``);` `  ``root.right.right.left = ``new` `Node(``6``);` `  ``root.right.right.right = ``new` `Node(``7``);` `  ``System.out.println(maxNonLeafNodesSum(root));` `}    ` `}`   `// This code is contributed by sanjeev2552`

## 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`

## C#

 `// C# implementation of ` `// the above approach ` `using` `System; ` `using` `System.Collections; `   `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;` `  `  `  ``public` `Node(``int` `data) ` `  ``{` `    ``this``.data = data;` `    ``this``.left = ``this``.right = ``null``;` `  ``}` `};` ` `  `// Function to return the maximum ` `// sum of non-leaf nodes  at any ` `// level in tree using level ` `// order traversal` `static` `int` `maxNonLeafNodesSum(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 a node ` `      ``// from queue` `      ``Node temp = (Node)q.Dequeue();` ` `  `      ``// 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.Enqueue(temp.left);` `      ``if` `(temp.right != ``null``)` `        ``q.Enqueue(temp.right);` `    ``}` ` `  `    ``// Update the maximum sum ` `    ``// of leaf nodes value` `    ``result = max(sum, result);` `  ``}` `  ``return` `result;` `}` ` `  `static` `int` `max(``int` `sum, ` `               ``int` `result) ` `{` `  ``if` `(sum > result)` `    ``return` `sum;` `  `  `  ``return` `result;` `}` ` `  `// Driver code` `public` `static` `void` `Main(``string``[] args) ` `{` `  ``Node root = ``new` `Node(1);` `  ``root.left = ``new` `Node(2);` `  ``root.right = ``new` `Node(3);` `  ``root.left.left = ``new` `Node(4);` `  ``root.left.right = ``new` `Node(5);` `  ``root.right.right = ``new` `Node(8);` `  ``root.right.right.left = ``new` `Node(6);` `  ``root.right.right.right = ``new` `Node(7);` `  ``Console.Write(maxNonLeafNodesSum(root));` `}    ` `}`   `// This code is contributed by rutvik_56`

## Javascript

 ``

Output:

`8`

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

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