# Sum of nodes in the left view of the given binary tree

Given a binary tree, the task is to find the sum of the nodes which are visible in the left view. The left view of a binary tree is the set of nodes visible when the tree is viewed from the left.
Examples:

```Input:
1
/  \
2    3
/ \    \
4   5    6
Output: 7
1 + 2 + 4 = 7

Input:
1
/  \
2      3
\
4
\
5
\
6
Output: 18```

Approach: The problem can be solved using simple recursive traversal. We can keep track of the level of a node by passing a parameter to all the recursive calls. The idea is to keep track of the maximum level also and traverse the tree in a manner that the left subtree is visited before the right subtree. Whenever a node whose level is more than the maximum level so far is encountered, add the value of the node to the sum because it is the first node in its level (Note that the left subtree is traversed before the right subtree).
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `class` `Node {` `public``:` `    ``int` `data;` `    ``Node *left, *right;` `};`   `// A utility function to create` `// a new Binary Tree Node` `Node* newNode(``int` `item)` `{` `    ``Node* temp = ``new` `Node();` `    ``temp->data = item;` `    ``temp->left = temp->right = NULL;` `    ``return` `temp;` `}`   `// Recursive function to find the sum of nodes` `// of the left view of the given binary tree` `void` `sumLeftViewUtil(Node* root, ``int` `level, ``int``* max_level, ``int``* sum)` `{` `    ``// Base Case` `    ``if` `(root == NULL)` `        ``return``;`   `    ``// If this is the first Node of its level` `    ``if` `(*max_level < level) {` `        ``*sum += root->data;` `        ``*max_level = level;` `    ``}`   `    ``// Recur for left and right subtrees` `    ``sumLeftViewUtil(root->left, level + 1, max_level, sum);` `    ``sumLeftViewUtil(root->right, level + 1, max_level, sum);` `}`   `// A wrapper over sumLeftViewUtil()` `void` `sumLeftView(Node* root)` `{` `    ``int` `max_level = 0;` `    ``int` `sum = 0;` `    ``sumLeftViewUtil(root, 1, &max_level, &sum);` `    ``cout << sum;` `}`   `// Driver code` `int` `main()` `{` `    ``Node* root = newNode(12);` `    ``root->left = newNode(10);` `    ``root->right = newNode(30);` `    ``root->right->left = newNode(25);` `    ``root->right->right = newNode(40);`   `    ``sumLeftView(root);`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach`   `// Class for a node of the tree` `class` `Node {` `    ``int` `data;` `    ``Node left, right;`   `    ``public` `Node(``int` `item)` `    ``{` `        ``data = item;` `        ``left = right = ``null``;` `    ``}` `}`   `class` `BinaryTree {` `    ``Node root;` `    ``static` `int` `max_level = ``0``;` `    ``static` `int` `sum = ``0``;`   `    ``// Recursive function to find the sum of nodes` `    ``// of the left view of the given binary tree` `    ``void` `sumLeftViewUtil(Node node, ``int` `level)` `    ``{` `        ``// Base Case` `        ``if` `(node == ``null``)` `            ``return``;`   `        ``// If this is the first node of its level` `        ``if` `(max_level < level) {` `            ``sum += node.data;` `            ``max_level = level;` `        ``}`   `        ``// Recur for left and right subtrees` `        ``sumLeftViewUtil(node.left, level + ``1``);` `        ``sumLeftViewUtil(node.right, level + ``1``);` `    ``}`   `    ``// A wrapper over sumLeftViewUtil()` `    ``void` `sumLeftView()` `    ``{`   `        ``sumLeftViewUtil(root, ``1``);` `        ``System.out.print(sum);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``BinaryTree tree = ``new` `BinaryTree();` `        ``tree.root = ``new` `Node(``12``);` `        ``tree.root.left = ``new` `Node(``10``);` `        ``tree.root.right = ``new` `Node(``30``);` `        ``tree.root.right.left = ``new` `Node(``25``);` `        ``tree.root.right.right = ``new` `Node(``40``);`   `        ``tree.sumLeftView();` `    ``}` `}`

## Python3

 `# Python3 implementation of the approach`   `# A binary tree node ` `class` `Node: `   `    ``# Constructor to create a new node ` `    ``def` `__init__(``self``, data): ` `        ``self``.data ``=` `data ` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None`     `# Recursive function to find the sum of nodes ` `# of the left view of the given binary tree` `def` `sumLeftViewUtil(root, level, max_level, ``sum``): ` `    `  `    ``# Base Case ` `    ``if` `root ``is` `None``: ` `        ``return`   `    ``# If this is the first node of its level ` `    ``if` `(max_level[``0``] < level): ` `        ``sum``[``0``]``+``=` `root.data ` `        ``max_level[``0``] ``=` `level `   `    ``# Recur for left and right subtree ` `    ``sumLeftViewUtil(root.left, level ``+` `1``, max_level, ``sum``) ` `    ``sumLeftViewUtil(root.right, level ``+` `1``, max_level, ``sum``) `     `# A wrapper over sumLeftViewUtil() ` `def` `sumLeftView(root): ` `    ``max_level ``=` `[``0``] ` `    ``sum` `=``[``0``]` `    ``sumLeftViewUtil(root, ``1``, max_level, ``sum``) ` `    ``print``(``sum``[``0``])`     `# Driver code` `root ``=` `Node(``12``) ` `root.left ``=` `Node(``10``) ` `root.right ``=` `Node(``20``) ` `root.right.left ``=` `Node(``25``) ` `root.right.right ``=` `Node(``40``) ` `sumLeftView(root)`

## C#

 `// C# implementation of the approach` `using` `System;`   `// Class for a node of the tree` `public` `class` `Node {` `    ``public` `int` `data;` `    ``public` `Node left, right;`   `    ``public` `Node(``int` `item)` `    ``{` `        ``data = item;` `        ``left = right = ``null``;` `    ``}` `}`   `public` `class` `BinaryTree {` `    ``public` `Node root;` `    ``public` `static` `int` `max_level = 0;` `    ``public` `static` `int` `sum = 0;`   `    ``// Recursive function to find the sum of nodes` `    ``// of the left view of the given binary tree` `    ``public` `virtual` `void` `leftViewUtil(Node node, ``int` `level)` `    ``{` `        ``// Base Case` `        ``if` `(node == ``null``) {` `            ``return``;` `        ``}`   `        ``// If this is the first node of its level` `        ``if` `(max_level < level) {` `            ``sum += node.data;` `            ``max_level = level;` `        ``}`   `        ``// Recur for left and right subtrees` `        ``leftViewUtil(node.left, level + 1);` `        ``leftViewUtil(node.right, level + 1);` `    ``}`   `    ``// A wrapper over leftViewUtil()` `    ``public` `virtual` `void` `leftView()` `    ``{` `        ``leftViewUtil(root, 1);` `        ``Console.Write(sum);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main(``string``[] args)` `    ``{` `        ``BinaryTree tree = ``new` `BinaryTree();` `        ``tree.root = ``new` `Node(12);` `        ``tree.root.left = ``new` `Node(10);` `        ``tree.root.right = ``new` `Node(30);` `        ``tree.root.right.left = ``new` `Node(25);` `        ``tree.root.right.right = ``new` `Node(40);`   `        ``tree.leftView();` `    ``}` `}`

## Javascript

 ``

Output

`47`

Time Complexity: O(N) where N is number of nodes in a given binary tree

Auxiliary Space: O(N)

Iterative Approach(Using Queue):
Follow the below steps  to solve the above problem:

• Simply traverse the whole binary tree in level Order Traversal with the help of Queue data structure.
• At each level keep track of the sum and add first node data of each level in sum.
• Finally, print the sum.
Below is the implementation of the above approach:

## C++

 `// THIS CODE IS CONTRIBUTED BY KIRTI AGARWAL(KIRTIAGARWAL23121999)` `// C++ Program for the above approach` `#include` `using` `namespace` `std;`   `struct` `Node{` `    ``int` `data;` `    ``Node* left;` `    ``Node* right;` `    ``Node(``int` `data){` `        ``this``->data = data;` `        ``this``->left = ``this``->right = NULL;` `    ``}` `};`   `// A utility function to create a new node` `struct` `Node* newNode(``int` `data){` `    ``return` `new` `Node(data);` `}`   `void` `sumLeftView(Node* root){` `    ``if``(root == NULL) ``return``;` `    `  `    ``queue q;` `    ``q.push(root);` `    `  `    ``int` `sum = 0;` `    ``while``(!q.empty()){` `        ``int` `n = q.size();` `        ``for``(``int` `i = 0; idata;` `            `  `            ``if``(front_node->left) q.push(front_node->left);` `            ``if``(front_node->right) q.push(front_node->right);` `        ``}` `    ``}` `    `  `    ``cout<left = newNode(10);` `    ``root->right = newNode(30);` `    ``root->right->left = newNode(25);` `    ``root->right->right = newNode(40);` `    `  `    ``sumLeftView(root);` `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.util.*;`   `class` `Node {` `    ``int` `data;` `    ``Node left, right;`   `    ``Node(``int` `data) {` `        ``this``.data = data;` `        ``left = right = ``null``;` `    ``}` `}`   `class` `Main {` `    ``static` `Node newNode(``int` `data) {` `        ``return` `new` `Node(data);` `    ``}`   `    ``static` `void` `sumLeftView(Node root) {` `        ``if` `(root == ``null``)` `            ``return``;`   `        ``Queue q = ``new` `LinkedList();` `        ``q.add(root);`   `        ``int` `sum = ``0``;` `        ``while` `(!q.isEmpty()) {` `            ``int` `n = q.size();` `            ``for` `(``int` `i = ``0``; i < n; i++) {` `                ``Node front_node = q.poll();` `                ``if` `(i == ``0``)` `                    ``sum = sum + front_node.data;`   `                ``if` `(front_node.left != ``null``)` `                    ``q.add(front_node.left);` `                ``if` `(front_node.right != ``null``)` `                    ``q.add(front_node.right);` `            ``}` `        ``}`   `        ``System.out.println(sum);` `    ``}`   `    ``// driver program to test above function` `    ``public` `static` `void` `main(String[] args) {` `        ``Node root = newNode(``12``);` `        ``root.left = newNode(``10``);` `        ``root.right = newNode(``30``);` `        ``root.right.left = newNode(``25``);` `        ``root.right.right = newNode(``40``);`   `        ``sumLeftView(root);` `    ``}` `}`

## Python3

 `# Python3 Program for the above approach` `from` `queue ``import` `Queue`   `class` `Node:` `    ``def` `__init__(``self``, data):` `        ``self``.data ``=` `data` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None`   `def` `sumLeftView(root):` `    ``if` `not` `root: ``return` `    `  `    ``q ``=` `Queue()` `    ``q.put(root)` `    `  `    ``sum` `=` `0` `    ``while` `not` `q.empty():` `        ``n ``=` `q.qsize()` `        ``for` `i ``in` `range``(n):` `            ``front_node ``=` `q.get()` `            ``if` `i ``=``=` `0``:` `                ``sum` `+``=` `front_node.data` `                `  `            ``if` `front_node.left:` `                ``q.put(front_node.left)` `            ``if` `front_node.right:` `                ``q.put(front_node.right)` `    `  `    ``print``(``sum``)`   `# driver program to test above function` `if` `__name__ ``=``=` `'__main__'``:` `    ``root ``=` `Node(``12``)` `    ``root.left ``=` `Node(``10``)` `    ``root.right ``=` `Node(``30``)` `    ``root.right.left ``=` `Node(``25``)` `    ``root.right.right ``=` `Node(``40``)` `    `  `    ``sumLeftView(root)`

## Javascript

 `// JavaScript program for the above approach` `class Node {` `constructor(data) {` `this``.data = data;` `this``.left = ``null``;` `this``.right = ``null``;` `}` `}`   `function` `sumLeftView(root) {` `if` `(root === ``null``)` `return``;` `let q = [];` `q.push(root);`   `let sum = 0;` `while` `(q.length > 0) {` `    ``let n = q.length;` `    ``for` `(let i = 0; i < n; i++) {` `        ``let front_node = q.shift();` `        ``if` `(i === 0)` `            ``sum = sum + front_node.data;`   `        ``if` `(front_node.left !== ``null``)` `            ``q.push(front_node.left);` `        ``if` `(front_node.right !== ``null``)` `            ``q.push(front_node.right);` `    ``}` `}`   `console.log(sum);` `}`   `// driver program to test above function` `function` `main() {` `let root = ``new` `Node(12);` `root.left = ``new` `Node(10);` `root.right = ``new` `Node(30);` `root.right.left = ``new` `Node(25);` `root.right.right = ``new` `Node(40);` `sumLeftView(root);` `}`   `main();` `// This code is contributed by Saroj Kumar Mandal`

## C#

 `// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;`   `class` `Node {` `    ``public` `int` `data;` `    ``public` `Node left, right;`   `    ``public` `Node(``int` `data) {` `        ``this``.data = data;` `        ``left = right = ``null``;` `    ``}` `}`   `class` `GFG {` `    ``static` `Node newNode(``int` `data) {` `        ``return` `new` `Node(data);` `    ``}`   `    ``static` `void` `sumLeftView(Node root) {` `        ``if` `(root == ``null``)` `            ``return``;`   `        ``Queue q = ``new` `Queue();` `        ``q.Enqueue(root);`   `        ``int` `sum = 0;` `        ``while` `(q.Count != 0) {` `            ``int` `n = q.Count;` `            ``for` `(``int` `i = 0; i < n; i++) {` `                ``Node front_node = q.Dequeue();` `                ``if` `(i == 0)` `                    ``sum = sum + front_node.data;`   `                ``if` `(front_node.left != ``null``)` `                    ``q.Enqueue(front_node.left);` `                ``if` `(front_node.right != ``null``)` `                    ``q.Enqueue(front_node.right);` `            ``}` `        ``}`   `        ``Console.WriteLine(sum);` `    ``}`   `    ``// driver program to test above function` `    ``public` `static` `void` `Main(``string``[] args) {` `        ``Node root = newNode(12);` `        ``root.left = newNode(10);` `        ``root.right = newNode(30);` `        ``root.right.left = newNode(25);` `        ``root.right.right = newNode(40);`   `        ``sumLeftView(root);` `    ``}` `}`

Output

`47`

Time Complexity: O(N) where N is the number of nodes in given binary tree.
Auxiliary Space: O(N) due to queue data structure.

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!