# Find the sum of leafs at maximum level

Given a binary tree containing n nodes. The task is to find the sum of all the leaf nodes present at maximum level.

Examples:

```Input:
1
/   \
2     3
/  \   /  \
4   5   6   7
/     \
8       9

Output: 17
Leaf nodes 8 and 9 are at maximum level.
Their sum = (8 + 9) = 17.

Input:
5
/   \
8     13
/
4

Output: 4
```

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

Approach: Perform iterative level order traversal using queue and find the sum of the nodes at each level and if there are no children for every node at the current level mark this level as the maximum level. Sum of all the leaf nodes at maximum level will be the required answer.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Structure of a node of binary tree ` `struct` `Node { ` `    ``int` `data; ` `    ``Node *left, *right; ` `}; ` ` `  `// Function to get a new node ` `Node* getNode(``int` `data) ` `{ ` `    ``// Allocate space ` `    ``Node* newNode = (Node*)``malloc``(``sizeof``(Node)); ` ` `  `    ``// Put in the data ` `    ``newNode->data = data; ` `    ``newNode->left = newNode->right = NULL; ` `    ``return` `newNode; ` `} ` ` `  `// Function to return the sum of the ` `// leaf nodes at maximum level ` `int` `sumOfLeafNodesAtMaxLevel(Node* root) ` `{ ` `    ``// If tree is empty ` `    ``if` `(!root) ` `        ``return` `0; ` ` `  `    ``// If there is only one node ` `    ``if` `(!root->left && !root->right) ` `        ``return` `root->data; ` ` `  `    ``// Queue used for level order traversal ` `    ``queue q; ` `    ``int` `leafSum = 0; ` `    ``bool` `f = 0; ` ` `  `    ``// Push root node in the queue 'q' ` `    ``q.push(root); ` ` `  `    ``while` `(``true``) { ` ` `  `        ``// Count number of nodes in the ` `        ``// current level ` `        ``int` `nc = q.size(); ` ` `  `        ``// If the queue is empty, it means that the ` `        ``// last processed level was the maximum level ` `        ``if` `(nc == 0) ` `            ``return` `leafSum; ` ` `  `        ``// Initialize leafSum for current level ` `        ``leafSum = 0; ` ` `  `        ``// Traverse the current level nodes ` `        ``while` `(nc--) { ` ` `  `            ``// Get front element from 'q' ` `            ``Node* top = q.front(); ` `            ``q.pop(); ` ` `  `            ``// If it is a leaf node ` `            ``if` `(!top->left && !top->right) { ` ` `  `                ``// Accumulate data to 'sum' ` `                ``leafSum += top->data; ` `            ``} ` `            ``else` `{ ` ` `  `                ``// If top's left and right child ` `                ``// exist then push them to 'q' ` `                ``if` `(top->left) ` `                    ``q.push(top->left); ` `                ``if` `(top->right) ` `                    ``q.push(top->right); ` `            ``} ` `        ``} ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``// Binary tree creation ` `    ``Node* root = getNode(1); ` `    ``root->left = getNode(2); ` `    ``root->right = getNode(3); ` `    ``root->left->left = getNode(4); ` `    ``root->left->right = getNode(5); ` `    ``root->right->left = getNode(6); ` `    ``root->right->right = getNode(7); ` `    ``root->left->right->left = getNode(8); ` `    ``root->right->left->right = getNode(9); ` ` `  `    ``cout << sumOfLeafNodesAtMaxLevel(root); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG { ` ` `  `    ``// Structure of a node of binary tree ` `    ``static` `class` `Node { ` `        ``int` `data; ` `        ``Node left, right; ` `    ``}; ` ` `  `    ``// Function to get a new node ` `    ``static` `Node getNode(``int` `data) ` `    ``{ ` `        ``// Allocate space ` `        ``Node newNode = ``new` `Node(); ` ` `  `        ``// Put in the data ` `        ``newNode.data = data; ` `        ``newNode.left = newNode.right = ``null``; ` `        ``return` `newNode; ` `    ``} ` ` `  `    ``// Function to return the sum of the ` `    ``// leaf nodes at maximum level ` `    ``static` `int` `sumOfLeafNodesAtMaxLevel(Node root) ` `    ``{ ` `        ``// If tree is empty ` `        ``if` `(root == ``null``) ` `            ``return` `0``; ` ` `  `        ``// If there is only one node ` `        ``if` `(root.left == ``null` `&& root.right == ``null``) ` `            ``return` `root.data; ` ` `  `        ``// Queue used for level order traversal ` `        ``Queue q = ``new` `LinkedList<>(); ` `        ``int` `leafSum = ``0``; ` `        ``boolean` `f = ``false``; ` ` `  `        ``// Push root node in the queue 'q' ` `        ``q.add(root); ` ` `  `        ``while` `(``true``) { ` ` `  `            ``// Count number of nodes in the ` `            ``// current level ` `            ``int` `nc = q.size(); ` ` `  `            ``// If the queue is empty, it means that the ` `            ``// last processed level was the maximum level ` `            ``if` `(nc == ``0``) ` `                ``return` `leafSum; ` ` `  `            ``// Initialize leafSum for current level ` `            ``leafSum = ``0``; ` ` `  `            ``// Traverse the current level nodes ` `            ``while` `(nc-- > ``0``) { ` ` `  `                ``// Get front element from 'q' ` `                ``Node top = q.peek(); ` `                ``q.remove(); ` ` `  `                ``// If it is a leaf node ` `                ``if` `(top.left == ``null` `&& top.right == ``null``) { ` ` `  `                    ``// Accumulate data to 'sum' ` `                    ``leafSum += top.data; ` `                ``} ` `                ``else` `{ ` ` `  `                    ``// If top's left and right child ` `                    ``// exist then push them to 'q' ` `                    ``if` `(top.left != ``null``) ` `                        ``q.add(top.left); ` `                    ``if` `(top.right != ``null``) ` `                        ``q.add(top.right); ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``// Binary tree creation ` `        ``Node root = getNode(``1``); ` `        ``root.left = getNode(``2``); ` `        ``root.right = getNode(``3``); ` `        ``root.left.left = getNode(``4``); ` `        ``root.left.right = getNode(``5``); ` `        ``root.right.left = getNode(``6``); ` `        ``root.right.right = getNode(``7``); ` `        ``root.left.right.left = getNode(``8``); ` `        ``root.right.left.right = getNode(``9``); ` ` `  `        ``System.out.print(sumOfLeafNodesAtMaxLevel(root)); ` `    ``} ` `} ` ` `  `// This code is contributed by Arnab Kundu `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG { ` ` `  `    ``// Structure of a node of binary tree ` `    ``public` `class` `Node { ` `        ``public` `int` `data; ` `        ``public` `Node left, right; ` `    ``}; ` ` `  `    ``// Function to get a new node ` `    ``static` `Node getNode(``int` `data) ` `    ``{ ` `        ``// Allocate space ` `        ``Node newNode = ``new` `Node(); ` ` `  `        ``// Put in the data ` `        ``newNode.data = data; ` `        ``newNode.left = newNode.right = ``null``; ` `        ``return` `newNode; ` `    ``} ` ` `  `    ``// Function to return the sum of the ` `    ``// leaf nodes at maximum level ` `    ``static` `int` `sumOfLeafNodesAtMaxLevel(Node root) ` `    ``{ ` `        ``// If tree is empty ` `        ``if` `(root == ``null``) ` `            ``return` `0; ` ` `  `        ``// If there is only one node ` `        ``if` `(root.left == ``null` `&& root.right == ``null``) ` `            ``return` `root.data; ` ` `  `        ``// Queue used for level order traversal ` `        ``Queue q = ``new` `Queue(); ` `        ``int` `leafSum = 0; ` ` `  `        ``// Push root node in the queue 'q' ` `        ``q.Enqueue(root); ` ` `  `        ``while` `(``true``) { ` ` `  `            ``// Count number of nodes in the ` `            ``// current level ` `            ``int` `nc = q.Count; ` ` `  `            ``// If the queue is empty, it means that the ` `            ``// last processed level was the maximum level ` `            ``if` `(nc == 0) ` `                ``return` `leafSum; ` ` `  `            ``// Initialize leafSum for current level ` `            ``leafSum = 0; ` ` `  `            ``// Traverse the current level nodes ` `            ``while` `(nc-- > 0) { ` ` `  `                ``// Get front element from 'q' ` `                ``Node top = q.Peek(); ` `                ``q.Dequeue(); ` ` `  `                ``// If it is a leaf node ` `                ``if` `(top.left == ``null` `&& top.right == ``null``) { ` ` `  `                    ``// Accumulate data to 'sum' ` `                    ``leafSum += top.data; ` `                ``} ` `                ``else` `{ ` ` `  `                    ``// If top's left and right child ` `                    ``// exist then push them to 'q' ` `                    ``if` `(top.left != ``null``) ` `                        ``q.Enqueue(top.left); ` `                    ``if` `(top.right != ``null``) ` `                        ``q.Enqueue(top.right); ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``// Binary tree creation ` `        ``Node root = getNode(1); ` `        ``root.left = getNode(2); ` `        ``root.right = getNode(3); ` `        ``root.left.left = getNode(4); ` `        ``root.left.right = getNode(5); ` `        ``root.right.left = getNode(6); ` `        ``root.right.right = getNode(7); ` `        ``root.left.right.left = getNode(8); ` `        ``root.right.left.right = getNode(9); ` ` `  `        ``Console.Write(sumOfLeafNodesAtMaxLevel(root)); ` `    ``} ` `} ` ` `  `// This code contributed by Rajput-Ji `

Output:

```17
```

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

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