Sum of nodes in the right 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 right view. The right view of a binary tree is the set of nodes visible when the tree is viewed from the right.

Examples:

Input:
       1
      /  \
     2    3
    / \    \
   4   5    6
Output: 10
1 + 3 + 6 = 10

Input:
       1
      /  \
    2      3
      \   
        4  
          \
            5
             \
               6
Output: 19

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 right subtree is visited before the left 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 last node in its level (Note that the right subtree is traversed before the left subtree).



Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 right view of the given binary tree
void sumRightViewUtil(Node* root, int level,
                      int* max_level, int* sum)
{
    // Base Case
    if (root == NULL)
        return;
  
    // If this is the last Node of its level
    if (*max_level < level) {
        *sum += root->data;
        *max_level = level;
    }
  
    // Recur for left and right subtrees
    sumRightViewUtil(root->right, level + 1, max_level, sum);
    sumRightViewUtil(root->left, level + 1, max_level, sum);
}
  
// A wrapper over sumRightViewUtil()
void sumRightView(Node* root)
{
    int max_level = 0;
    int sum = 0;
    sumRightViewUtil(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);
  
    sumRightView(root);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 right view of the given binary tree
    void sumRightViewUtil(Node node, int level)
    {
        // Base Case
        if (node == null)
            return;
  
        // If this is the last node of its level
        if (max_level < level) {
            sum += node.data;
            max_level = level;
        }
  
        // Recur for left and right subtrees
        sumRightViewUtil(node.right, level + 1);
        sumRightViewUtil(node.left, level + 1);
    }
  
    // A wrapper over sumRightViewUtil()
    void sumRightView()
    {
  
        sumRightViewUtil(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.sumRightView();
    }
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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 right view of the given binary tree
def sumRightViewUtil(root, level, max_level, sum): 
      
    # Base Case 
    if root is None
        return
  
    # If this is the last node of its level 
    if (max_level[0] < level): 
        sum[0]+= root.data 
        max_level[0] = level 
  
    # Recur for left and right subtree 
    sumRightViewUtil(root.right, level + 1, max_level, sum
    sumRightViewUtil(root.left, level + 1, max_level, sum
      
  
# A wrapper over sumRightViewUtil() 
def sumRightView(root): 
    max_level = [0
    sum =[0]
    sumRightViewUtil(root, 1, max_level, sum
    print(sum[0])
  
  
# Driver code
root = Node(12
root.left = Node(10
root.right = Node(30
root.right.left = Node(25
root.right.right = Node(40
sumRightView(root)

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 right view of the given binary tree
    public virtual void sumrightViewUtil(Node node, int level)
    {
        // Base Case
        if (node == null) {
            return;
        }
  
        // If this is the last node of its level
        if (max_level < level) {
            sum += node.data;
            max_level = level;
        }
  
        // Recur for left and right subtrees
        sumrightViewUtil(node.right, level + 1);
        sumrightViewUtil(node.left, level + 1);
    }
  
    // A wrapper over sumrightViewUtil()
    public virtual void sumrightView()
    {
        sumrightViewUtil(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.sumrightView();
    }
}

chevron_right


Output:

82


My Personal Notes arrow_drop_up

Small things always make you to think big

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.




Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.