# Find maximum level sum in Binary Tree

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

Examples:

`Input :               4                    /   \                   2    -5                  / \    /\                -1   3 -2  6Output: 6Explanation :Sum of all nodes of 0'th level is 4Sum of all nodes of 1'th level is -3Sum of all nodes of 0'th level is 6Hence maximum sum is 6Input :          1               /   \             2      3           /  \      \          4    5      8                    /   \                   6     7  Output :  17`

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

Below is the implementation of the above idea:

## C++

 `// 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); ``} `` ` `// 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); `` ` `    ``/*   Constructed Binary tree is: ``                 ``1 ``               ``/   \ ``             ``2      3 ``           ``/  \      \ ``          ``4    5      8 ``                    ``/   \ ``                   ``6     7    */``    ``cout << ``"Maximum level sum is "` `<< maxLevelSum(root) ``         ``<< endl; ``    ``return` `0; ``}`

## Java

 `// A queue based Java program to find maximum  ``// sum of a level in Binary Tree ``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 find the maximum  ``// sum of a level in tree ``// using level order traversal ``static` `int` `maxLevelSum(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 = ``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 an node from queue ``            ``Node temp = q.poll(); `` ` `            ``// 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.add(temp.left); ``            ``if` `(temp.right != ``null``) ``                ``q.add(temp.right); ``        ``} `` ` `        ``// Update the maximum node ``        ``// count value ``        ``result = Math.max(sum, result); ``    ``} ``    ``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``); ``     ` `    ``/*   Constructed Binary tree is: ``                 ``1 ``               ``/   \ ``             ``2      3 ``           ``/  \      \ ``          ``4    5      8 ``                    ``/   \ ``                   ``6     7    */``    ``System.out.println(``"Maximum level sum is "` `+ ``                        ``maxLevelSum(root)); ``} ``} `` ` `// This code is contributed by sanjeev2552`

## Python3

 `# A queue based Python3 program to find ``# maximum sum of a level in Binary Tree ``from` `collections ``import` `deque `` ` `# A binary tree node has data, pointer ``# to left child and a pointer to right  ``# child  ``class` `Node: ``     ` `    ``def` `__init__(``self``, key): ``         ` `        ``self``.data ``=` `key ``        ``self``.left ``=` `None``        ``self``.right ``=` `None`` ` `# Function to find the maximum sum  ``# of a level in tree ``# using level order traversal ``def` `maxLevelSum(root): ``     ` `    ``# Base case ``    ``if` `(root ``=``=` `None``): ``        ``return` `0`` ` `    ``# Initialize result ``    ``result ``=` `root.data ``     ` `    ``# Do Level order traversal keeping ``    ``# track of number ``    ``# of nodes at every level. ``    ``q ``=` `deque() ``    ``q.append(root) ``     ` `    ``while` `(``len``(q) > ``0``): ``         ` `        ``# Get the size of queue when the  ``        ``# level order traversal for one  ``        ``# level finishes ``        ``count ``=` `len``(q) `` ` `        ``# Iterate for all the nodes in ``        ``# the queue currently ``        ``sum` `=` `0``        ``while` `(count > ``0``): ``             ` `            ``# Dequeue an node from queue ``            ``temp ``=` `q.popleft() `` ` `            ``# Add this node's value to current sum. ``            ``sum` `=` `sum` `+` `temp.data `` ` `            ``# Enqueue left and right children of ``            ``# dequeued node ``            ``if` `(temp.left !``=` `None``): ``                ``q.append(temp.left) ``            ``if` `(temp.right !``=` `None``): ``                ``q.append(temp.right) ``                 ` `            ``count ``-``=` `1`    ` ` `        ``# Update the maximum node count 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``) `` ` `    ``# Constructed Binary tree is: ``    ``#              1 ``    ``#            /   \ ``    ``#          2      3 ``    ``#        /  \      \ ``    ``#       4    5      8 ``    ``#                 /   \ ``    ``#                6     7     ``    ``print``(``"Maximum level sum is"``, maxLevelSum(root)) `` ` `# This code is contributed by mohit kumar 29`

## C#

 `// A queue based C# program to find maximum  ``// sum of a level in Binary Tree ``using` `System; ``using` `System.Collections.Generic; ``class` `GFG ``{ `` ` `  ``// A binary tree node has data, pointer ``  ``// to left child and a pointer to right ``  ``// child ``  ``public``    ``class` `Node  ``    ``{ ``      ``public``        ``int` `data; ``      ``public``        ``Node left, right; `` ` `      ``public` `Node(``int` `data) ``      ``{ ``        ``this``.data = data; ``        ``this``.left = ``this``.right = ``null``; ``      ``} ``    ``}; `` ` `  ``// Function to find the maximum  ``  ``// sum of a level in tree ``  ``// using level order traversal ``  ``static` `int` `maxLevelSum(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 = ``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 an node from queue ``        ``Node temp = q.Dequeue(); `` ` `        ``// 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.Enqueue(temp.left); ``        ``if` `(temp.right != ``null``) ``          ``q.Enqueue(temp.right); ``      ``} `` ` `      ``// Update the maximum node ``      ``// count value ``      ``result = Math.Max(sum, result); ``    ``} ``    ``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); `` ` `    ``/*   Constructed Binary tree is: ``                 ``1 ``               ``/   \ ``             ``2      3 ``           ``/  \      \ ``          ``4    5      8 ``                    ``/   \ ``                   ``6     7    */``    ``Console.WriteLine(``"Maximum level sum is "` `+ ``                      ``maxLevelSum(root)); ``  ``} ``} `` ` `// This code is contributed by gauravrajput1 `

## Javascript

 ``

Output
```Maximum level sum is 17

```

Complexity Analysis:

Time Complexity: O(N) where N is the total number of nodes in the tree.
In level order traversal, every node of the tree is processed once, and hence the complexity due to the level order traversal is O(N) if there are total N nodes in the tree. Also, while processing every node, we are maintaining the sum at each level, however, this does not affect the overall time complexity. Therefore, the time complexity is O(N).

Auxiliary Space: O(w) where w is the maximum width of the tree.
In level order traversal, a queue is maintained whose maximum size at any moment can go up to the maximum width of the binary tree.

Using Recursion Without Queue:-

• We will use recursion and do any dfs of the tree.
• As in dfs we will move downwards of the tree so while moving we will take care of the level of the tree
• We will add the node value to the current level of the tree
• In the end we will return the maximum sum from all level

Implementation:-

• We will start traversal by level 0 that is from root
• From root we will move downwards using recursion and while moving we will increase the level of the tree by +1.
• We will take a unordered_map to store the sum of the current level
• In the end we will return the maximum value from the map.

## C++

 `// 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 get sum or each level ``void` `dfs(Node* root,``int` `level,unordered_map<``int``,``int``> &mm) ``{ ``  ``//base condition ``  ``if``(!root)``return``; ``   ` `  ``//adding root value to its level sum ``  ``mm[level]+=root->data; ``   ` `  ``//increasing level ``  ``level++; ``   ` `  ``//moving left ``  ``dfs(root->left,level,mm); ``   ` `  ``//moving right ``  ``dfs(root->right,level,mm); ``} ``// 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; ``   ` `      ``//map to store sum of each level ``      ``unordered_map<``int``,``int``> mm; ``   ` `      ``//calling function ``      ``dfs(root,0,mm); ``   ` `      ``//variable to store answer ``    ``int` `result = INT_MIN;     ``   ` `      ``//iterating over map ``      ``for``(``auto` `x:mm)result = max(x.second,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); `` ` `    ``/*   Constructed Binary tree is: ``                 ``1 ``               ``/   \ ``             ``2      3 ``           ``/  \      \ ``          ``4    5      8 ``                    ``/   \ ``                   ``6     7    */``    ``cout << ``"Maximum level sum is "` `<< maxLevelSum(root) ``         ``<< endl; ``    ``return` `0; ``} `` ` `//code contributed by shubhamrajput6156`

## Java

 `import` `java.util.*; `` ` `// A binary tree node has data, a pointer to the left child, ``// and a pointer to the right child ``class` `Node { ``    ``int` `data; ``    ``Node left, right; `` ` `    ``// Constructor to create a new node ``    ``Node(``int` `data) { ``        ``this``.data = data; ``        ``left = right = ``null``; ``    ``} ``} `` ` `class` `GFG { ``    ``// Function to get the sum for each level ``    ``static` `void` `dfs(Node root, ``int` `level, Map mm) { ``        ``// Base condition ``        ``if` `(root == ``null``) ``return``; `` ` `        ``// Adding the root value to its level sum ``        ``mm.put(level, mm.getOrDefault(level, ``0``) + root.data); `` ` `        ``// Increasing level ``        ``level++; `` ` `        ``// Moving left ``        ``dfs(root.left, level, mm); `` ` `        ``// Moving right ``        ``dfs(root.right, level, mm); ``    ``} `` ` `    ``// Function to find the maximum sum of a level in  ``  ``// the tree using level order traversal ``    ``static` `int` `maxLevelSum(Node root) { ``        ``// Base case ``        ``if` `(root == ``null``) ``return` `0``; `` ` `        ``// Map to store the sum of each level ``        ``Map mm = ``new` `HashMap<>(); `` ` `        ``// Calling the dfs function ``        ``dfs(root, ``0``, mm); `` ` `        ``// Variable to store the answer ``        ``int` `result = Integer.MIN_VALUE; `` ` `        ``// Iterating over the map ``        ``for` `(Map.Entry entry : mm.entrySet()) { ``            ``result = Math.max(entry.getValue(), result); ``        ``} `` ` `        ``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``); `` ` `        ``// Constructed binary tree is: ``        ``//        1 ``        ``//      /   \ ``        ``//     2     3 ``        ``//    / \     \ ``        ``//   4   5     8 ``        ``//          /   \ ``        ``//         6     7 `` ` `        ``System.out.println(``"Maximum level sum is "` `+ maxLevelSum(root)); ``    ``} ``} `

## Python3

 `# A binary tree node has data, left and right pointers ``class` `Node: ``    ``def` `__init__(``self``, data): ``        ``self``.data ``=` `data ``        ``self``.left ``=` `None``        ``self``.right ``=` `None`` ` `# Function to get sum of each level ``def` `dfs(root, level, mm): ``    ``# Base condition ``    ``if` `not` `root: ``        ``return`` ` `    ``# Adding root value to its level sum ``    ``mm[level] ``=` `mm.get(level, ``0``) ``+` `root.data `` ` `    ``# Increasing level ``    ``level ``+``=` `1`` ` `    ``# Moving left ``    ``dfs(root.left, level, mm) `` ` `    ``# Moving right ``    ``dfs(root.right, level, mm) `` ` `# Function to find the maximum sum of a level in the tree ``def` `maxLevelSum(root): ``    ``# Base case ``    ``if` `not` `root: ``        ``return` `0`` ` `    ``# Map to store the sum of each level ``    ``mm ``=` `{} `` ` `    ``# Calling the function to calculate the sum of each level ``    ``dfs(root, ``0``, mm) `` ` `    ``# Variable to store the answer ``    ``result ``=` `float``(``'-inf'``) `` ` `    ``# Iterating over the map to find the maximum sum ``    ``for` `val ``in` `mm.values(): ``        ``result ``=` `max``(result, val) `` ` `    ``return` `result `` ` `# Helper function to allocate a new node with the given data and NULL left and right pointers ``def` `newNode(data): ``    ``node ``=` `Node(data) ``    ``return` `node `` ` `# Driver code ``if` `__name__ ``=``=` `"__main__"``: ``    ``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 ``    ``"""``    ``print``(``"Maximum level sum is"``, maxLevelSum(root)) `

## C#

 `using` `System; ``using` `System.Collections.Generic; `` ` `class` `Node ``{ ``    ``public` `int` `data; ``    ``public` `Node left, right; `` ` `    ``public` `Node(``int` `item) ``    ``{ ``        ``data = item; ``        ``left = right = ``null``; ``    ``} ``} `` ` `public` `class` `MainClass ``{ ``    ``// Function to get sum of each level ``    ``static` `void` `DFS(Node root, ``int` `level, Dictionary<``int``, ``int``> mm) ``    ``{ ``        ``// Base condition ``        ``if` `(root == ``null``) ``            ``return``; `` ` `        ``// Adding root value to its level sum ``        ``if` `(mm.ContainsKey(level)) ``            ``mm[level] += root.data; ``        ``else``            ``mm[level] = root.data; `` ` `        ``// Increasing level ``        ``level++; `` ` `        ``// Moving left ``        ``DFS(root.left, level, mm); `` ` `        ``// Moving right ``        ``DFS(root.right, level, mm); ``    ``} `` ` `    ``// Function to find the maximum sum of a level in the tree ``    ``static` `int` `MaxLevelSum(Node root) ``    ``{ ``        ``// Base case ``        ``if` `(root == ``null``) ``            ``return` `0; `` ` `        ``// Dictionary to store the sum of each level ``        ``Dictionary<``int``, ``int``> mm = ``new` `Dictionary<``int``, ``int``>(); `` ` `        ``// Calling the function to calculate the sum of each level ``        ``DFS(root, 0, mm); `` ` `        ``// Variable to store the answer ``        ``int` `result = ``int``.MinValue; `` ` `        ``// Iterating over the dictionary to find the maximum sum ``        ``foreach` `(``var` `val ``in` `mm.Values) ``        ``{ ``            ``result = Math.Max(result, val); ``        ``} `` ` `        ``return` `result; ``    ``} `` ` `    ``// Helper function to allocate a new node with the given data and NULL left and right pointers ``    ``static` `Node NewNode(``int` `data) ``    ``{ ``        ``Node node = ``new` `Node(data); ``        ``return` `node; ``    ``} `` ` `    ``// Driver code ``    ``public` `static` `void` `Main(``string``[] args) ``    ``{ ``        ``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 */``        ``Console.WriteLine(``"Maximum level sum is "` `+ MaxLevelSum(root)); ``    ``} ``} `

## Javascript

 `class Node { ``    ``constructor(data) { ``        ``this``.data = data; ``        ``this``.left = ``null``; ``        ``this``.right = ``null``; ``    ``} ``} `` ` `// Function to get sum of each level ``function` `dfs(root, level, mm) { ``    ``// Base condition ``    ``if` `(!root) { ``        ``return``; ``    ``} `` ` `    ``// Adding root value to its level sum ``    ``mm[level] = (mm[level] || 0) + root.data; `` ` `    ``// Increasing level ``    ``level++; `` ` `    ``// Moving left ``    ``dfs(root.left, level, mm); `` ` `    ``// Moving right ``    ``dfs(root.right, level, mm); ``} `` ` `// Function to find the maximum sum of a level in the tree ``function` `maxLevelSum(root) { ``    ``// Base case ``    ``if` `(!root) { ``        ``return` `0; ``    ``} `` ` `    ``// Object to store the sum of each level ``    ``const mm = {}; `` ` `    ``// Calling the function to calculate the sum of each level ``    ``dfs(root, 0, mm); `` ` `    ``// Variable to store the answer ``    ``let result = Number.MIN_SAFE_INTEGER; `` ` `    ``// Iterating over the object to find the maximum sum ``    ``for` `(let val of Object.values(mm)) { ``        ``result = Math.max(result, val); ``    ``} `` ` `    ``return` `result; ``} `` ` `// Helper function to allocate a new node with the given data  ``// and NULL left and right pointers ``function` `newNode(data) { ``    ``return` `new` `Node(data); ``} `` ` `// Driver code `` ` `    ``const 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 ``    ``*/``    ``console.log(``"Maximum level sum is"``, maxLevelSum(root)); `

Output:- Maximum level sum is 17

Time Complexity:- O(N)

Space Complexity:- O(H) where H is height of tree

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