# Sum of nodes on the longest path from root to leaf node

• Difficulty Level : Medium
• Last Updated : 12 Jan, 2023

Given a binary tree containing n nodes. The problem is to find the sum of all nodes on the longest path from root to leaf node. If two or more paths compete for the longest path, then the path having maximum sum of nodes is being considered.
Examples:

```Input : Binary tree:
4
/ \
2   5
/ \ / \
7  1 2  3
/
6
Output : 13

4
/ \
2   5
/ \ / \
7  1 2  3
/
6

The highlighted nodes (4, 2, 1, 6) above are
part of the longest root to leaf path having
sum = (4 + 2 + 1 + 6) = 13```

Approach: Recursively find the length and sum of nodes of each root to leaf path and accordingly update the maximum sum.
Algorithm:

```sumOfLongRootToLeafPath(root, sum, len, maxLen, maxSum)
if root == NULL
if maxLen < len
maxLen = len
maxSum = sum
else if maxLen == len && maxSum is less than sum
maxSum = sum
return

sumOfLongRootToLeafPath(root-left, sum + root-data,
len + 1, maxLen, maxSum)
sumOfLongRootToLeafPath(root-right, sum + root-data,
len + 1, maxLen, maxSum)

sumOfLongRootToLeafPathUtil(root)
if (root == NULL)
return 0

Declare maxSum = Minimum Integer
Declare maxLen = 0
sumOfLongRootToLeafPath(root, 0, 0, maxLen, maxSum)
return maxSum```

## C++

 `// C++ implementation to find the sum of nodes``// on the longest path from root to leaf node``#include ` `using` `namespace` `std;` `// Node of a binary tree``struct` `Node {``    ``int` `data;``    ``Node* left, *right;``};` `// function to get a new node``Node* getNode(``int` `data)``{``    ``// allocate memory for the node``    ``Node* newNode = (Node*)``malloc``(``sizeof``(Node));` `    ``// put in the data``    ``newNode->data = data;``    ``newNode->left = newNode->right = NULL;``    ``return` `newNode;``}` `// function to find the sum of nodes on the``// longest path from root to leaf node``void` `sumOfLongRootToLeafPath(Node* root, ``int` `sum,``                      ``int` `len, ``int``& maxLen, ``int``& maxSum)``{``    ``// if true, then we have traversed a``    ``// root to leaf path``    ``if` `(!root) {``        ``// update maximum length and maximum sum``        ``// according to the given conditions``        ``if` `(maxLen < len) {``            ``maxLen = len;``            ``maxSum = sum;``        ``} ``else` `if` `(maxLen == len && maxSum < sum)``            ``maxSum = sum;``        ``return``;``    ``}` `    ``// recur for left subtree``    ``sumOfLongRootToLeafPath(root->left, sum + root->data,``                            ``len + 1, maxLen, maxSum);` `    ``// recur for right subtree``    ``sumOfLongRootToLeafPath(root->right, sum + root->data,``                            ``len + 1, maxLen, maxSum);``}` `// utility function to find the sum of nodes on``// the longest path from root to leaf node``int` `sumOfLongRootToLeafPathUtil(Node* root)``{``    ``// if tree is NULL, then sum is 0``    ``if` `(!root)``        ``return` `0;` `    ``int` `maxSum = INT_MIN, maxLen = 0;` `    ``// finding the maximum sum 'maxSum' for the``    ``// maximum length root to leaf path``    ``sumOfLongRootToLeafPath(root, 0, 0, maxLen, maxSum);` `    ``// required maximum sum``    ``return` `maxSum;``}` `// Driver program to test above``int` `main()``{``    ``// binary tree formation``    ``Node* root = getNode(4);         ``/*        4        */``    ``root->left = getNode(2);         ``/*       / \       */``    ``root->right = getNode(5);        ``/*      2   5      */``    ``root->left->left = getNode(7);   ``/*     / \ / \     */``    ``root->left->right = getNode(1);  ``/*    7  1 2  3    */``    ``root->right->left = getNode(2);  ``/*      /          */``    ``root->right->right = getNode(3); ``/*     6           */``    ``root->left->right->left = getNode(6);` `    ``cout << ``"Sum = "``         ``<< sumOfLongRootToLeafPathUtil(root);` `    ``return` `0;``}`

## Java

 `// Java implementation to find the sum of nodes``// on the longest path from root to leaf node``public` `class` `GFG``{                       ``    ``// Node of a binary tree``    ``static` `class` `Node {``        ``int` `data;``        ``Node left, right;``        ` `        ``Node(``int` `data){``            ``this``.data = data;``            ``left = ``null``;``            ``right = ``null``;``        ``}``    ``}``    ``static` `int` `maxLen;``    ``static` `int` `maxSum;``    ` `    ``// function to find the sum of nodes on the``    ``// longest path from root to leaf node``    ``static` `void` `sumOfLongRootToLeafPath(Node root, ``int` `sum,``                                         ``int` `len)``    ``{``        ``// if true, then we have traversed a``        ``// root to leaf path``        ``if` `(root == ``null``) {``            ``// update maximum length and maximum sum``            ``// according to the given conditions``            ``if` `(maxLen < len) {``                ``maxLen = len;``                ``maxSum = sum;``            ``} ``else` `if` `(maxLen == len && maxSum < sum)``                ``maxSum = sum;``            ``return``;``        ``}``        ` `        ` `        ``// recur for left subtree``        ``sumOfLongRootToLeafPath(root.left, sum + root.data,``                                ``len + ``1``);``        ` `        ``sumOfLongRootToLeafPath(root.right, sum + root.data,``                                ``len + ``1``);``        ` `    ``}``     ` `    ``// utility function to find the sum of nodes on``    ``// the longest path from root to leaf node``    ``static` `int` `sumOfLongRootToLeafPathUtil(Node root)``    ``{``        ``// if tree is NULL, then sum is 0``        ``if` `(root == ``null``)``            ``return` `0``;``     ` `        ``maxSum = Integer.MIN_VALUE;``        ``maxLen = ``0``;``     ` `        ``// finding the maximum sum 'maxSum' for the``        ``// maximum length root to leaf path``        ``sumOfLongRootToLeafPath(root, ``0``, ``0``);``     ` `        ``// required maximum sum``        ``return` `maxSum;``    ``}``     ` `    ``// Driver program to test above``    ``public` `static` `void` `main(String args[])``    ``{``        ``// binary tree formation``        ``Node root = ``new` `Node(``4``);         ``/*        4        */``        ``root.left = ``new` `Node(``2``);         ``/*       / \       */``        ``root.right = ``new` `Node(``5``);        ``/*      2   5      */``        ``root.left.left = ``new` `Node(``7``);    ``/*     / \ / \     */``        ``root.left.right = ``new` `Node(``1``);   ``/*    7  1 2  3    */``        ``root.right.left = ``new` `Node(``2``);   ``/*      /          */``        ``root.right.right = ``new` `Node(``3``);  ``/*     6           */``        ``root.left.right.left = ``new` `Node(``6``);``     ` `        ``System.out.println( ``"Sum = "``             ``+ sumOfLongRootToLeafPathUtil(root));``    ``}``}``// This code is contributed by Sumit Ghosh`

## Python3

 `# Python3 implementation to find the ``# Sum of nodes on the longest path``# from root to leaf nodes` `# function to get a new node``class` `getNode:``    ``def` `__init__(``self``, data):` `        ``# put in the data``        ``self``.data ``=` `data``        ``self``.left ``=` `self``.right ``=` `None` `# function to find the Sum of nodes on the``# longest path from root to leaf node``def` `SumOfLongRootToLeafPath(root, ``Sum``, ``Len``,``                            ``maxLen, maxSum):``                                ` `    ``# if true, then we have traversed a``    ``# root to leaf path``    ``if` `(``not` `root):``        ` `        ``# update maximum Length and maximum Sum``        ``# according to the given conditions``        ``if` `(maxLen[``0``] < ``Len``):``            ``maxLen[``0``] ``=` `Len``            ``maxSum[``0``] ``=` `Sum``        ``else` `if` `(maxLen[``0``]``=``=` `Len` `and``              ``maxSum[``0``] < ``Sum``):``            ``maxSum[``0``] ``=` `Sum``        ``return` `    ``# recur for left subtree``    ``SumOfLongRootToLeafPath(root.left, ``Sum` `+` `root.data,``                            ``Len` `+` `1``, maxLen, maxSum)` `    ``# recur for right subtree``    ``SumOfLongRootToLeafPath(root.right, ``Sum` `+` `root.data,``                            ``Len` `+` `1``, maxLen, maxSum)` `# utility function to find the Sum of nodes on``# the longest path from root to leaf node``def` `SumOfLongRootToLeafPathUtil(root):``    ` `    ``# if tree is NULL, then Sum is 0``    ``if` `(``not` `root):``        ``return` `0` `    ``maxSum ``=` `[``-``999999999999``]``    ``maxLen ``=` `[``0``]` `    ``# finding the maximum Sum 'maxSum' for``    ``# the maximum Length root to leaf path``    ``SumOfLongRootToLeafPath(root, ``0``, ``0``,``                            ``maxLen, maxSum)` `    ``# required maximum Sum``    ``return` `maxSum[``0``]` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# binary tree formation``    ``root ``=` `getNode(``4``)         ``#     4    ``    ``root.left ``=` `getNode(``2``)         ``#     / \    ``    ``root.right ``=` `getNode(``5``)     ``#     2 5    ``    ``root.left.left ``=` `getNode(``7``) ``#     / \ / \    ``    ``root.left.right ``=` `getNode(``1``) ``# 7 1 2 3``    ``root.right.left ``=` `getNode(``2``) ``#     /        ``    ``root.right.right ``=` `getNode(``3``) ``#     6        ``    ``root.left.right.left ``=` `getNode(``6``)` `    ``print``(``"Sum = "``, SumOfLongRootToLeafPathUtil(root))``    ` `# This code is contributed by PranchalK`

## C#

 `using` `System;` `// c# implementation to find the sum of nodes``// on the longest path from root to leaf node``public` `class` `GFG``{``    ``// Node of a binary tree``    ``public` `class` `Node``    ``{``        ``public` `int` `data;``        ``public` `Node left, right;` `        ``public` `Node(``int` `data)``        ``{``            ``this``.data = data;``            ``left = ``null``;``            ``right = ``null``;``        ``}``    ``}``    ``public` `static` `int` `maxLen;``    ``public` `static` `int` `maxSum;` `    ``// function to find the sum of nodes on the``    ``// longest path from root to leaf node``    ``public` `static` `void` `sumOfLongRootToLeafPath(Node root, ``int` `sum, ``int` `len)``    ``{``        ``// if true, then we have traversed a``        ``// root to leaf path``        ``if` `(root == ``null``)``        ``{``            ``// update maximum length and maximum sum``            ``// according to the given conditions``            ``if` `(maxLen < len)``            ``{``                ``maxLen = len;``                ``maxSum = sum;``            ``}``            ``else` `if` `(maxLen == len && maxSum < sum)``            ``{``                ``maxSum = sum;``            ``}``            ``return``;``        ``}`  `        ``// recur for left subtree``        ``sumOfLongRootToLeafPath(root.left, sum + root.data, len + 1);` `        ``sumOfLongRootToLeafPath(root.right, sum + root.data, len + 1);` `    ``}` `    ``// utility function to find the sum of nodes on``    ``// the longest path from root to leaf node``    ``public` `static` `int` `sumOfLongRootToLeafPathUtil(Node root)``    ``{``        ``// if tree is NULL, then sum is 0``        ``if` `(root == ``null``)``        ``{``            ``return` `0;``        ``}` `        ``maxSum = ``int``.MinValue;``        ``maxLen = 0;` `        ``// finding the maximum sum 'maxSum' for the``        ``// maximum length root to leaf path``        ``sumOfLongRootToLeafPath(root, 0, 0);` `        ``// required maximum sum``        ``return` `maxSum;``    ``}` `    ``// Driver program to test above``    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``// binary tree formation``        ``Node root = ``new` `Node(4); ``//        4``        ``root.left = ``new` `Node(2); ``//       / \``        ``root.right = ``new` `Node(5); ``//      2   5``        ``root.left.left = ``new` `Node(7); ``//     / \ / \``        ``root.left.right = ``new` `Node(1); ``//    7  1 2  3``        ``root.right.left = ``new` `Node(2); ``//      /``        ``root.right.right = ``new` `Node(3); ``//     6``        ``root.left.right.left = ``new` `Node(6);` `        ``Console.WriteLine(``"Sum = "` `+ sumOfLongRootToLeafPathUtil(root));``    ``}``}` `  ``// This code is contributed by Shrikant13`

## Javascript

 ``

Output

`Sum = 13`

Time Complexity: O(n)

Another Approach: Using level order traversal

1. Create a structure containing the current Node, level and sum in the path.
2. Push the root element with level 0 and sum as the root’s data.
3. Pop the front element and update the maximum level sum and maximum level if needed.
4. Push the left and right nodes if exists.
5. Do the same for all the nodes in tree.

## C++

 `#include ``using` `namespace` `std;` `//Building a tree node having left and right pointers set to null initially``struct` `Node``{``  ``Node* left;``  ``Node* right;``  ``int` `data;``  ``//constructor to set the data of the newly created tree node``  ``Node(``int` `element){``     ``data = element;``     ``this``->left = nullptr;``     ``this``->right = nullptr;``  ``}``};` `int` `longestPathLeaf(Node* root){``  ` `  ``/* structure to store current Node,it's level and sum in the path*/``  ``struct` `Element{``    ``Node* data;``    ``int` `level;``    ``int` `sum;``  ``};``  ` `  ``/*``    ``maxSumLevel stores maximum sum so far in the path``    ``maxLevel stores maximum level so far``  ``*/``  ``int` `maxSumLevel = root->data,maxLevel = 0;` `  ``/* queue to implement level order traversal */``  ` `  ``list que;``  ``Element e;``  ` `  ``/* Each element variable stores the current Node, it's level, sum in the path */` `  ``e.data = root;``  ``e.level = 0;``  ``e.sum = root->data;``  ` `  ``/* push the root element*/``  ``que.push_back(e);``  ` `  ``/* do level order traversal on the tree*/``  ``while``(!que.empty()){` `     ``Element front = que.front();``     ``Node* curr = front.data;``     ``que.pop_front();``     ` `     ``/* if the level of current front element is greater than the maxLevel so far then update maxSum*/``     ``if``(front.level > maxLevel){``        ``maxSumLevel = front.sum;``        ``maxLevel = front.level;``     ``}``     ``/* if another path competes then update if the sum is greater than the previous path of same height*/``     ``else` `if``(front.level == maxLevel && front.sum > maxSumLevel)``        ``maxSumLevel = front.sum;` `     ``/* push the left element if exists*/` `     ``if``(curr->left){``        ``e.data = curr->left;``        ``e.sum = e.data->data;``        ``e.sum +=  front.sum;``        ``e.level = front.level+1;``        ``que.push_back(e);``     ``}``     ``/*push the right element if exists*/``     ``if``(curr->right){``        ``e.data = curr->right;``        ``e.sum = e.data->data;``        ``e.sum +=  front.sum;``        ``e.level = front.level+1;``        ``que.push_back(e);``     ``}``  ``}` `  ``/*return the answer*/``  ``return` `maxSumLevel;``}``//Helper function``int` `main() {``  ` `  ``Node* root = ``new` `Node(4);        ``  ``root->left = ``new` `Node(2);        ``  ``root->right = ``new` `Node(5);       ``  ``root->left->left = ``new` `Node(7);  ``  ``root->left->right = ``new` `Node(1); ``  ``root->right->left = ``new` `Node(2);``  ``root->right->right = ``new` `Node(3);``  ``root->left->right->left = ``new` `Node(6);``  ` `  ``cout << longestPathLeaf(root) << ``"\n"``;``    ` `  ``return` `0;``}`

## Java

 `// Java Code to find sum of nodes on the longest path from``// root to leaf node using level order traversal``import` `java.util.*;` `// Building a tree node having left and right pointers set``// to null initially``class` `Main {` `    ``static` `class` `Node {``        ``Node left;``        ``Node right;``        ``int` `data;``        ``// constructor to set the data of the newly created``        ``// tree node``        ``Node(``int` `element)``        ``{``            ``data = element;``            ``this``.left = ``null``;``            ``this``.right = ``null``;``        ``}``    ``}` `    ``/* structure to store current Node,it's level and sum in``     ``* the path*/``    ``static` `class` `Element {``        ``Node data;``        ``int` `level;``        ``int` `sum;``    ``}` `    ``public` `static` `int` `longestPathLeaf(Node root)``    ``{``        ``/*``          ``maxSumLevel stores maximum sum so far in the path``          ``maxLevel stores maximum level so far``        ``*/``        ``int` `maxSumLevel = root.data;``        ``int` `maxLevel = ``0``;` `        ``/* queue to implement level order traversal */``        ``Queue que = ``new` `LinkedList<>();``        ``Element e = ``new` `Element();` `        ``/* Each element variable stores the current Node,``         ``* it's level, sum in the path */` `        ``e.data = root;``        ``e.level = ``0``;``        ``e.sum = root.data;` `        ``/* push the root element*/``        ``que.add(e);` `        ``/* do level order traversal on the tree*/``        ``while` `(!que.isEmpty()) {``            ``Element front = que.poll();``            ``Node curr = front.data;` `            ``/* if the level of current front element is``             ``* greater than the maxLevel so far then update``             ``* maxSum*/``            ``if` `(front.level > maxLevel) {``                ``maxSumLevel = front.sum;``                ``maxLevel = front.level;``            ``}` `            ``/* if another path competes then update if the``             ``* sum is greater than the previous path of same``             ``* height*/``            ``else` `if` `(front.level == maxLevel``                     ``&& front.sum > maxSumLevel) {``                ``maxSumLevel = front.sum;``            ``}` `            ``/* push the left element if exists*/``            ``if` `(curr.left != ``null``) {``                ``e = ``new` `Element();``                ``e.data = curr.left;``                ``e.sum = e.data.data;``                ``e.sum += front.sum;``                ``e.level = front.level + ``1``;``                ``que.add(e);``            ``}` `            ``/*push the right element if exists*/``            ``if` `(curr.right != ``null``) {``                ``e = ``new` `Element();``                ``e.data = curr.right;``                ``e.sum = e.data.data;``                ``e.sum += front.sum;``                ``e.level = front.level + ``1``;``                ``que.add(e);``            ``}``        ``}``        ``/*return the answer*/``        ``return` `maxSumLevel;``    ``}``    ``// Helper function``    ``public` `static` `void` `main(String[] args)``    ``{` `        ``Node root = ``new` `Node(``4``);``        ``root.left = ``new` `Node(``2``);``        ``root.right = ``new` `Node(``5``);``        ``root.left.left = ``new` `Node(``7``);``        ``root.left.right = ``new` `Node(``1``);``        ``root.right.left = ``new` `Node(``2``);``        ``root.right.right = ``new` `Node(``3``);``        ``root.left.right.left = ``new` `Node(``6``);` `        ``System.out.println(longestPathLeaf(root));``    ``}``}` `// This code is contributed by Tapesh(tapeshdua420)`

## Python3

 `# Python Code to find sum of nodes on the longest path from root to leaf node``# using level order traversal` `# Building a tree node having left and right pointers set to null initially``class` `Node:``    ``def` `__init__(``self``, element):``        ``self``.data ``=` `element``        ``self``.left ``=` `None``        ``self``.right ``=` `None` `# To store current Node,it's level and sum in the path``class` `Element:``    ``def` `__init__(``self``, data, level, ``sum``):``        ``self``.data ``=` `data``        ``self``.level ``=` `level``        ``self``.``sum` `=` `sum`  `class` `Solution:``    ``def` `longestPathLeaf(``self``, root):` `        ``# maxSumLevel stores maximum sum so far in the path``        ``# maxLevel stores maximum level so far` `        ``maxSumLevel ``=` `root.data``        ``maxLevel ``=` `0` `        ``# queue to implement level order traversal``        ``que ``=` `[]` `        ``# Each element variable stores the current Node, it's level, sum in the path``        ``e ``=` `Element(root, ``0``, root.data)` `        ``# push the root element``        ``que.append(e)` `        ``# do level order traversal on the tree``        ``while` `len``(que) !``=` `0``:` `            ``front ``=` `que[``0``]``            ``curr ``=` `front.data``            ``del` `que[``0``]` `           ``# if the level of current front element is greater than the maxLevel so far then update maxSum``            ``if` `front.level > maxLevel:``                ``maxSumLevel ``=` `front.``sum``                ``maxLevel ``=` `front.level` `                ``# if another path competes then update if the sum is greater than the previous path of same height``            ``elif` `front.level ``=``=` `maxLevel ``and` `front.``sum` `> maxSumLevel:``                ``maxSumLevel ``=` `front.``sum` `                ``# push the left element if exists``            ``if` `curr.left !``=` `None``:``                ``e ``=` `Element(curr.left, front.level``+``1``, curr.left.data``+``front.``sum``)``                ``que.append(e)` `                ``# push the right element if exists``            ``if` `curr.right !``=` `None``:``                ``e ``=` `Element(curr.right, front.level``+``1``,``                            ``curr.right.data``+``front.``sum``)``                ``que.append(e)` `        ``# return the answer``        ``return` `maxSumLevel`  `# Helper function``if` `__name__ ``=``=` `'__main__'``:``    ``s ``=` `Solution()``    ``root ``=` `Node(``4``)``    ``root.left ``=` `Node(``2``)``    ``root.right ``=` `Node(``5``)``    ``root.left.left ``=` `Node(``7``)``    ``root.left.right ``=` `Node(``1``)``    ``root.right.left ``=` `Node(``2``)``    ``root.right.right ``=` `Node(``3``)``    ``root.left.right.left ``=` `Node(``6``)` `    ``print``(s.longestPathLeaf(root))` `# This code is contributed by Tapesh(tapeshdua420)`

## C#

 `// C# program to find sum of nodes on the longest path from``// root to leaf node using level order traversal``using` `System;``using` `System.Collections;` `// Building a tree node having left and right pointers set``// to null initially``class` `Node {` `  ``public` `Node left;``  ``public` `Node right;``  ``public` `int` `data;` `  ``// constructor to set the data of the newly created``  ``// tree node``  ``public` `Node(``int` `element)``  ``{` `    ``data = element;``    ``this``.left = ``null``;``    ``this``.right = ``null``;``  ``}``}``/* structure to store current Node,it's level and sum in`` ``* the path*/``class` `Element {` `  ``public` `Node data;``  ``public` `int` `level;``  ``public` `int` `sum;``}` `class` `GFG {``  ``public` `static` `int` `longestPathLeaf(Node root)``  ``{``    ``/*``          ``maxSumLevel stores maximum sum so far in the path``          ``maxLevel stores maximum level so far``        ``*/``    ``int` `maxSumLevel = root.data;``    ``int` `maxLevel = 0;` `    ``/* queue to implement level order traversal */``    ``Queue que = ``new` `Queue();``    ``Element e = ``new` `Element();` `    ``/* Each element variable stores the current Node,``         ``* it's level, sum in the path */` `    ``e.data = root;``    ``e.level = 0;``    ``e.sum = root.data;` `    ``/* push the root element*/``    ``que.Enqueue(e);` `    ``/* do level order traversal on the tree*/``    ``while` `(que.Count != 0) {``      ``dynamic front = que.Dequeue();``      ``Node curr = front.data;` `      ``/* if the level of current front element is``             ``* greater than the maxLevel so far then update``             ``* maxSum*/``      ``if` `(front.level > maxLevel) {``        ``maxSumLevel = front.sum;``        ``maxLevel = front.level;``      ``}` `      ``/* if another path competes then update if the``             ``* sum is greater than the previous path of same``             ``* height*/``      ``else` `if` `(front.level == maxLevel``               ``&& front.sum > maxSumLevel) {``        ``maxSumLevel = front.sum;``      ``}` `      ``/* push the left element if exists*/``      ``if` `(curr.left != ``null``) {``        ``e = ``new` `Element();``        ``e.data = curr.left;``        ``e.sum = e.data.data;``        ``e.sum += front.sum;``        ``e.level = front.level + 1;``        ``que.Enqueue(e);``      ``}` `      ``/*push the right element if exists*/``      ``if` `(curr.right != ``null``) {``        ``e = ``new` `Element();``        ``e.data = curr.right;``        ``e.sum = e.data.data;``        ``e.sum += front.sum;``        ``e.level = front.level + 1;``        ``que.Enqueue(e);``      ``}``    ``}``    ``/*return the answer*/``    ``return` `maxSumLevel;``  ``}``  ``// Helper function``  ``public` `static` `void` `Main(``string``[] args)``  ``{` `    ``Node root = ``new` `Node(4);``    ``root.left = ``new` `Node(2);``    ``root.right = ``new` `Node(5);``    ``root.left.left = ``new` `Node(7);``    ``root.left.right = ``new` `Node(1);``    ``root.right.left = ``new` `Node(2);``    ``root.right.right = ``new` `Node(3);``    ``root.left.right.left = ``new` `Node(6);` `    ``Console.WriteLine(longestPathLeaf(root));``  ``}``}` `// This code is contributed by Tapesh(tapeshdua420)`

## Javascript

 `// Javascript code to find sum of nodes on the longest path from root to leaf node``// using level order traversal` `class Node {``    ``constructor(element) {``        ``this``.data = element;``        ``this``.left = ``null``;``        ``this``.right = ``null``;``    ``}``}` `class Element {``    ``constructor(data, level, sum) {``        ``this``.data = data;``        ``this``.level = level;``        ``this``.sum = sum;``    ``}``}` `class Solution {``    ``longestPathLeaf(root) {` `        ``// maxSumLevel stores maximum sum so far in the path``        ``// maxLevel stores maximum level so far` `        ``let maxSumLevel = root.data;``        ``let maxLevel = 0;` `        ``// queue to implement level order traversal``        ``let que = [];` `        ``// Each element variable stores the current Node, it's level, sum in the path``        ``let e = ``new` `Element(root, 0, root.data);` `        ``// push the root element``        ``que.push(e);` `        ``// do level order traversal on the tree``        ``while` `(que.length !== 0) {` `            ``let front = que[0];``            ``let curr = front.data;``            ``que.shift();` `            ``// if the level of current front element is greater than the maxLevel so far then update maxSum``            ``if` `(front.level > maxLevel) {``                ``maxSumLevel = front.sum;``                ``maxLevel = front.level;``            ``}``            ``// if another path competes then update if the sum is greater than the previous path of same height``            ``else` `if` `(front.level === maxLevel && front.sum > maxSumLevel) {``                ``maxSumLevel = front.sum;``            ``}` `            ``// push the left element if exists``            ``if` `(curr.left !== ``null``) {``                ``e = ``new` `Element(curr.left, front.level + 1, curr.left.data + front.sum);``                ``que.push(e);``            ``}` `            ``// push the right element if exists``            ``if` `(curr.right !== ``null``) {``                ``e = ``new` `Element(curr.right, front.level + 1, curr.right.data + front.sum);``                ``que.push(e);``            ``}``        ``}` `        ``// return the answer``        ``return` `maxSumLevel;``    ``}``}` `// Helper function``const s = ``new` `Solution();``const root = ``new` `Node(4);``root.left = ``new` `Node(2);``root.right = ``new` `Node(5);``root.left.left = ``new` `Node(7);``root.left.right = ``new` `Node(1);``root.right.left = ``new` `Node(2);``root.right.right = ``new` `Node(3);``root.left.right.left = ``new` `Node(6);` `console.log(s.longestPathLeaf(root));`

Output

`13`

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

Contributed by Manjukrishna

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