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

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)

Auxiliary Space: O(h) where h is the height of the binary tree.

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

In this approach, we perform a BFS on the tree and keep track of the maximum depth and the sum of nodes at that depth. We can calculate the sum of longest bloodline by adding the maximum depth to the sum of nodes at that depth.

Follow the steps below to implement the above approach:

1. Initialize a variable max_sum to the minimum integer value.
2. Create an empty queue for BFS traversal.
3. Enqueue the root node to the queue.
4. While the queue is not empty, do the following:
a. Initialize a variable level_sum to 0.
b. Get the number of nodes in the current level.
c. For each node in the current level, do the following:
i. Dequeue a node from the queue.
ii. Add the node’s data to the level_sum.
iii. Enqueue the left child of the node to the queue if it exists.
iv. Enqueue the right child of the node to the queue if it exists.
d. Update the max_sum with the level_sum if it’s greater.
5. Return the max_sum

Below is the implementation of the above approach:

## C++

 `// C++ code to implement BFS approach` `#include ` `using` `namespace` `std;`   `// Define a struct to represent a node of the tree` `struct` `Node {` `    ``int` `data;` `    ``Node* left;` `    ``Node* right;` `};`   `// Function to create a new node with the given data` `Node* newNode(``int` `data) {` `    ``Node* node = ``new` `Node;` `    ``node->data = data;` `    ``node->left = NULL;` `    ``node->right = NULL;` `    ``return` `node;` `}`   `// Function to perform BFS on the tree and calculate the sum of longest bloodline` `int` `bfs(Node* root) {` `    ``int` `max_sum = INT_MIN;  ``// Initialize max_sum to the minimum integer value` `    ``queue q;  ``// Create an empty queue for BFS traversal` `    ``q.push(root);  ``// Enqueue the root node` `    ``while` `(!q.empty()) {` `        ``int` `level_sum = 0;  ``// Initialize the level_sum to 0` `        ``int` `n = q.size();  ``// Get the number of nodes in the current level` `        ``for` `(``int` `i = 0; i < n; i++) {` `            ``Node* node = q.front();  ``// Dequeue a node from the queue` `            ``q.pop();` `            ``level_sum += node->data;  ``// Add the node's data to the level_sum` `            ``if` `(node->left) q.push(node->left);  ``// Enqueue the left child of the node if it exists` `            ``if` `(node->right) q.push(node->right);  ``// Enqueue the right child of the node if it exists` `        ``}` `        ``max_sum = max(max_sum, level_sum);  ``// Update the max_sum with the level_sum if it's greater` `    ``}` `    ``return` `max_sum;  ``// Return the max_sum` `}`   `// Driver code ` `int` `main() {` `    ``// Create the tree` `    ``Node* root = newNode(4);` `    ``root->left = newNode(2);` `    ``root->right = newNode(5);` `    ``root->left->left = newNode(7);` `    ``root->left->right = newNode(1);` `    ``root->right->left = newNode(2);` `    ``root->right->right = newNode(3);` `    ``root->left->right->left = newNode(6);`   `    ``// Calculate the sum of longest bloodline using BFS` `    ``int` `sum = bfs(root);`   `    ``// Print the result` `    ``cout << sum << endl;`   `    ``return` `0;` `}` `// This code is contributed by Veerendra_Singh_Rajpoot`

## Python3

 `from` `collections ``import` `deque`   `# Define a class to represent a node of the tree` `class` `Node:` `    ``def` `__init__(``self``, data):` `        ``self``.data ``=` `data` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None`   `# Function to perform BFS on the tree and calculate the sum of longest bloodline` `def` `bfs(root):` `    ``max_sum ``=` `float``(``'-inf'``)  ``# Initialize max_sum to negative infinity` `    ``q ``=` `deque()  ``# Create an empty queue for BFS traversal` `    ``q.append(root)  ``# Enqueue the root node` `    ``while` `q:` `        ``level_sum ``=` `0`  `# Initialize the level_sum to 0` `        ``n ``=` `len``(q)  ``# Get the number of nodes in the current level` `        ``for` `i ``in` `range``(n):` `            ``node ``=` `q.popleft()  ``# Dequeue a node from the queue` `            ``level_sum ``+``=` `node.data  ``# Add the node's data to the level_sum` `            ``if` `node.left: q.append(node.left)  ``# Enqueue the left child of the node if it exists` `            ``if` `node.right: q.append(node.right)  ``# Enqueue the right child of the node if it exists` `        ``max_sum ``=` `max``(max_sum, level_sum)  ``# Update the max_sum with the level_sum if it's greater` `    ``return` `max_sum  ``# Return the max_sum`   `# Driver code ` `if` `__name__ ``=``=` `'__main__'``:` `    ``# Create the tree` `    ``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``)`   `    ``# Calculate the sum of longest bloodline using BFS` `    ``sum` `=` `bfs(root)`   `    ``# Print the result` `    ``print``(``sum``)`

## Javascript

 `// Define a class to represent a node of the tree` `class Node {` `    ``constructor(data) {` `        ``this``.data = data;` `        ``this``.left = ``null``;` `        ``this``.right = ``null``;` `    ``}` `}`   `// Function to perform BFS on the tree and calculate the sum of longest bloodline` `function` `bfs(root) {` `    ``let max_sum = Number.MIN_SAFE_INTEGER;  ``// Initialize max_sum to the minimum safe integer value` `    ``let q = [];  ``// Create an empty queue for BFS traversal` `    ``q.push(root);  ``// Enqueue the root node` `    ``while` `(q.length > 0) {` `        ``let level_sum = 0;  ``// Initialize the level_sum to 0` `        ``let n = q.length;  ``// Get the number of nodes in the current level` `        ``for` `(let i = 0; i < n; i++) {` `            ``let node = q.shift();  ``// Dequeue a node from the queue` `            ``level_sum += node.data;  ``// Add the node's data to the level_sum` `            ``if` `(node.left) q.push(node.left);  ``// Enqueue the left child of the node if it exists` `            ``if` `(node.right) q.push(node.right);  ``// Enqueue the right child of the node if it exists` `        ``}` `        ``max_sum = Math.max(max_sum, level_sum);  ``// Update the max_sum with the level_sum if it's greater` `    ``}` `    ``return` `max_sum;  ``// Return the max_sum` `}`   `// Create the tree` `let 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);`   `// Calculate the sum of longest bloodline using BFS` `let sum = bfs(root);`   `// Print the result` `console.log(sum);`

Output

`13`

Time Complexity: O(N), where N is the number of nodes in the tree.
Space Complexity: O(W), where W is the maximum width of the tree (queue size).

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.