# Find maximum level sum in Binary Tree

Given a Binary Tree having positive and negative nodes, the task is to find maximum sum level in it.

Examples:

```Input :               4
/   \
2    -5
/ \    /\
-1   3 -2  6
Output: 6
Explanation :
Sum of all nodes of 0'th level is 4
Sum of all nodes of 1'th level is -3
Sum of all nodes of 0'th level is 6
Hence maximum sum is 6

Input :          1
/   \
2      3
/  \      \
4    5      8
/   \
6     7
Output :  17
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

This problem is a variation of maximum width problem. The idea is to do level order traversal of tree. While doing traversal, process nodes of different level separately. For every level being processed, compute sum of nodes in the level and keep track of maximum sum.

 `// A queue based C++ program to find maximum sum ` `// of a level in Binary Tree ` `#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 find the maximum sum of a level in tree ` `// using level order traversal ` `int` `maxLevelSum(``struct` `Node * root) ` `{ ` `    ``// Base case ` `    ``if` `(root == NULL) ` `        ``return` `0; ` ` `  `    ``// Initialize result ` `    ``int` `result = root->data; ` ` `  `    ``// Do Level order traversal keeping track of 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 an node from queue ` `            ``Node *temp = q.front(); ` `            ``q.pop(); ` ` `  `            ``// Add this node's value to current sum. ` `            ``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 node count 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); ` `} ` ` `  `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); ` ` `  `    ``/*   Constructed Binary tree is: ` `                 ``1 ` `               ``/   \ ` `             ``2      3 ` `           ``/  \      \ ` `          ``4    5      8 ` `                    ``/   \ ` `                   ``6     7    */` `    ``cout << ``"Maximum level sum is "` `         ``<< maxLevelSum(root) << endl; ` `    ``return` `0; ` `} `

Output :

`Maximum level sum is 17`

Time Complexity : O(n)
Auxiliary Space : O(n)

This article is contributed by Shashank Mishra ( Gullu ). 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.