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Sum of nodes in the left view of the given binary tree
  • Last Updated : 20 Sep, 2019

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();
    }
}


Output:

47

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