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 <bits/stdc++.h>
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

<script>

// Javascript implementation of the approach

// Class for a node of the tree
class Node {

    constructor(item)
    {
        this.data = item;
        this.left = null;
        this.right = null;
    }
}

var root = null;
var max_level = 0;
var sum = 0;

// Recursive function to find the sum of nodes
// of the left view of the given binary tree
function leftViewUtil(node, 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()
function leftView()
{
    leftViewUtil(root, 1);
    document.write(sum);
}

// Driver code
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);
leftView();

// This code is contributed by rrrtnx.
</script>

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<bits/stdc++.h>
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<Node*> q;
    q.push(root);

    int sum = 0;
    while(!q.empty()){
        int n = q.size();
        for(int i = 0; i<n; i++){
            Node* front_node = q.front();
            q.pop();
            if(i == 0) sum = sum + front_node->data;

            if(front_node->left) q.push(front_node->left);
            if(front_node->right) q.push(front_node->right);
        }
    }

    cout<<sum<<endl;
}

// driver program to test above function
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 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<Node> q = new LinkedList<Node>();
        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<Node> q = new Queue<Node>();
        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.