Print Right View of a Binary Tree

Given a Binary Tree, print Right view of it. Right view of a Binary Tree is set of nodes visible when tree is visited from Right side. 

Right view of following tree is 1 3 7 8

          1
       /     \
     2        3
   /   \     /  \
  4     5   6    7
                  \
                   8

The problem can be solved using simple recursive traversal. We can keep track of level of a node by passing a parameter to all recursive calls. The idea is to keep track of maximum level also. And traverse the tree in a manner that right subtree is visited before left subtree. Whenever we see a node whose level is more than maximum level so far, we print the node because this is the last node in its level (Note that we traverse the right subtree before left subtree). Following is the implementation of this approach. 

C++

// C++ program to print right view of Binary Tree
#include <bits/stdc++.h>
using namespace std;

struct Node
{
    int data;
    struct Node *left, *right;
};

// A utility function to
// create a new Binary Tree Node
struct Node *newNode(int item)
{
    struct Node *temp = (struct Node *)malloc(
                          sizeof(struct Node));
    temp->data = item;
    temp->left = temp->right = NULL;
    return temp;
}

// Recursive function to print
// right view of a binary tree.
void rightViewUtil(struct Node *root,
                   int level, int *max_level)
{
    // Base Case
    if (root == NULL) return;

    // If this is the last Node of its level
    if (*max_level < level)
    {
        cout << root->data << "\t";
        *max_level = level;
    }

    // Recur for right subtree first,
    // then left subtree
    rightViewUtil(root->right, level + 1, max_level);
    rightViewUtil(root->left, level + 1, max_level);
}

// A wrapper over rightViewUtil()
void rightView(struct Node *root)
{
    int max_level = 0;
    rightViewUtil(root, 1, &max_level);
}

// Driver Code
int main()
{
    struct Node *root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
    root->right->right->right = newNode(8);

    rightView(root);

    return 0;
}

// This code is contributed by SHUBHAMSINGH10

C

// C program to print right view of Binary Tree
#include<stdio.h>
#include<stdlib.h>

struct Node
{
    int data;
    struct Node *left, *right;
};

// A utility function to create a new Binary Tree Node
struct Node *newNode(int item)
{
    struct Node *temp =  (struct Node *)malloc(sizeof(struct Node));
    temp->data = item;
    temp->left = temp->right = NULL;
    return temp;
}

// Recursive function to print right view of a binary tree.
void rightViewUtil(struct Node *root, int level, int *max_level)
{
    // Base Case
    if (root==NULL)  return;

    // If this is the last Node of its level
    if (*max_level < level)
    {
        printf("%d\t", root->data);
        *max_level = level;
    }

    // Recur for right subtree first, then left subtree
    rightViewUtil(root->right, level+1, max_level);
    rightViewUtil(root->left, level+1, max_level);
}

// A wrapper over rightViewUtil()
void rightView(struct Node *root)
{
    int max_level = 0;
    rightViewUtil(root, 1, &max_level);
}

// Driver Program to test above functions
int main()
{
    struct Node *root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
    root->right->left->right = newNode(8);

    rightView(root);

    return 0;
}

Java

// Java program to print right view of binary tree

// A binary tree node
class Node {

    int data;
    Node left, right;

    Node(int item) {
        data = item;
        left = right = null;
    }
}

// class to access maximum level by reference
class Max_level {

    int max_level;
}

class BinaryTree {

    Node root;
    Max_level max = new Max_level();

    // Recursive function to print right view of a binary tree.
    void rightViewUtil(Node node, int level, Max_level max_level) {

        // Base Case
        if (node == null)
            return;

        // If this is the last Node of its level
        if (max_level.max_level < level) {
            System.out.print(node.data + " ");
            max_level.max_level = level;
        }

        // Recur for right subtree first, then left subtree
        rightViewUtil(node.right, level + 1, max_level);
        rightViewUtil(node.left, level + 1, max_level);
    }

    void rightView()
    {
        rightView(root);
    }

    // A wrapper over rightViewUtil()
    void rightView(Node node) {

        rightViewUtil(node, 1, max);
    }

    // Driver program to test the above functions
    public static void main(String args[]) {
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(5);
        tree.root.right.left = new Node(6);
        tree.root.right.right = new Node(7);
        tree.root.right.left.right = new Node(8);

        tree.rightView();

        }
}

// This code has been contributed by Mayank Jaiswal

Python

# Python program to print right view of Binary Tree

# A binary tree node
class Node:
    # A constructor to create a new Binary tree Node
    def __init__(self, item):
        self.data = item
        self.left = None
        self.right = None

# Recursive function to print right view of Binary Tree
# used max_level as reference list ..only max_level[0]
# is helpful to us
def rightViewUtil(root, level, max_level):

    # Base Case
    if root is None:
        return

    # If this is the last node of its level
    if (max_level[0] < level):
        print "%d   " %(root.data),
        max_level[0] = level

    # Recur for right subtree first, then left subtree
    rightViewUtil(root.right, level+1, max_level)
    rightViewUtil(root.left, level+1, max_level)

def rightView(root):
    max_level = [0]
    rightViewUtil(root, 1, max_level)


# Driver program to test above function
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left = Node(6)
root.right.right = Node(7)
root.right.left.right = Node(8)

rightView(root)

# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

C#

using System;

// C# program to print right view of binary tree

// A binary tree node
public class Node
{

    public int data;
    public Node left, right;

    public Node(int item)
    {
        data = item;
        left = right = null;
    }
}

// class to access maximum level by reference
public class Max_level
{

    public int max_level;
}

public class BinaryTree
{

    public Node root;
    public Max_level max = new Max_level();

    // Recursive function to print right view of a binary tree.
    public virtual void rightViewUtil(Node node, int level,
                                        Max_level max_level)
    {

        // Base Case
        if (node == null)
        {
            return;
        }

        // If this is the last Node of its level
        if (max_level.max_level < level)
        {
            Console.Write(node.data + " ");
            max_level.max_level = level;
        }

        // Recur for right subtree first, then left subtree
        rightViewUtil(node.right, level + 1, max_level);
        rightViewUtil(node.left, level + 1, max_level);
    }

    public virtual void rightView()
    {
        rightView(root);
    }

    // A wrapper over rightViewUtil()
    public virtual void rightView(Node node)
    {

        rightViewUtil(node, 1, max);
    }

    // Driver program to test the above functions
    public static void Main(string[] args)
    {
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(5);
        tree.root.right.left = new Node(6);
        tree.root.right.right = new Node(7);
        tree.root.right.left.right = new Node(8);

        tree.rightView();

    }
}

// This code is contributed by Shrikant13

Javascript

<script>

    // JavaScript program to print
    // right view of binary tree

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

    let max_level = 0;

    let root;

    // Recursive function to print right view of a binary tree.
    function rightViewUtil(node, level) {

        // Base Case
        if (node == null)
            return;

        // If this is the last Node of its level
        if (max_level < level) {
            document.write(node.data + " ");
            max_level = level;
        }

        // Recur for right subtree first, then left subtree
        rightViewUtil(node.right, level + 1);
        rightViewUtil(node.left, level + 1);
    }

    function rightView()
    {
        rightview(root);
    }

    // A wrapper over rightViewUtil()
    function rightview(node) {

        rightViewUtil(node, 1);
    }

    root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(3);
    root.left.left = new Node(4);
    root.left.right = new Node(5);
    root.right.left = new Node(6);
    root.right.right = new Node(7);
    root.right.left.right = new Node(8);

    rightView();

</script>

Output

1    3    7    8    

Right view of Binary Tree using Queue
Time Complexity: The function does a simple traversal of the tree, so the complexity is O(n).

Auxiliary Space: O(n)

Method 2: In this method, level order traversal based solution is discussed. If we observe carefully, we will see that our main task is to print the right most node of every level. So, we will do a level order traversal on the tree and print the last node at every level. Below is the implementation of above approach:

C++

// C++ program to print left view of
// Binary Tree

#include <bits/stdc++.h>
using namespace std;

// A Binary Tree Node
struct Node {
    int data;
    struct Node *left, *right;
};

// Utility function to create a new tree node
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}

// function to print Right view of
// binary tree
void printRightView(Node* root)
{
    if (root == NULL)
        return;

    queue<Node*> q;
    q.push(root);

    while (!q.empty()) {
        // get number of nodes for each level
        int n = q.size();

        // traverse all the nodes of the current level
        while (n--) {

            Node* x = q.front();
            q.pop();

            // print the last node of each level
            if (n == 0) {
                cout << x->data << " ";
            }
            // if left child is not null push it into the
            // queue
            if (x->left)
                q.push(x->left);
            // if right child is not null push it into the
            // queue
            if (x->right)
                q.push(x->right);
        }
    }
}

// Driver code
int main()
{
    // Let's construct the tree as
    // shown in example

    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
    root->right->left->right = newNode(8);

    printRightView(root);
}

// This code is contributed by
// Snehasish Dhar

Java

// JAVA program to print right view of
// Binary Tree

import java.io.*;
import java.util.LinkedList;
import java.util.Queue;

// A Binary Tree Node
class Node {
    int data;
    Node left, right;
    public Node(int d)
    {
        data = d;
        left = right = null;
    }
}

class BinaryTree {
    Node root;

    // function to print Right view of
    // binary tree
    void rightView(Node root)
    {
        if (root == null) {
            return;
        }

        Queue<Node> q = new LinkedList<>();
        q.add(root);

        while (!q.isEmpty()) {

            // get number of nodes for each level
            int n = q.size();

            // traverse all the nodes of the current level
            for (int i = 0; i < n; i++) {
                Node curr = q.peek();
                q.remove();

                // print the last node of each level
                if (i == n - 1) {
                    System.out.print(curr.data);
                    System.out.print(" ");
                }

                // if left child is not null add it into
                // the
                // queue
                if (curr.left != null) {
                    q.add(curr.left);
                }

                // if right child is not null add it into
                // the
                // queue
                if (curr.right != null) {
                    q.add(curr.right);
                }
            }
        }
    }

    // Driver code
    public static void main(String[] args)
    {

        // Let's construct the tree as
        // shown in example
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(5);
        tree.root.right.left = new Node(6);
        tree.root.right.right = new Node(7);
        tree.root.right.left.right = new Node(8);

        tree.rightView(tree.root);
    }
}

// This code is contributed by Biswajit Rajak

Python3

# Python3 program to print right
# view of Binary Tree
from collections import deque

# A binary tree node
class Node:

    # A constructor to create a new
    # Binary tree Node
    def __init__(self, val):
        self.data = val
        self.left = None
        self.right = None

# Function to print Right view of
# binary tree
def rightView(root):

    if root is None:
        return

    q = deque()
    q.append(root)

    while q:

        # Get number of nodes for each level
        n = len(q)

        # Traverse all the nodes of the
        # current level

        while n > 0:
            n -= 1

            # Get the front node in the queue
            node = q.popleft()

            # Print the last node of each level
            if n == 0:
                print(node.data, end = " ")

            # If left child is not null push it
            # into the queue
            if node.left:
                q.append(node.left)

            # If right child is not null push
            # it into the queue
            if node.right:
                q.append(node.right)

# Driver code

# Let's construct the tree as
# shown in example
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left = Node(6)
root.right.right = Node(7)
root.right.left.right = Node(8)

rightView(root)

# This code is contributed by Pulkit Pansari

C#

// C# program to print right view of
// Binary Tree
using System;
using System.Collections.Generic;

// A Binary Tree Node
public  class Node {
  public    int data;
  public Node left, right;
  public Node(int d)
  {
    data = d;
    left = right = null;
  }
}


public class BinaryTree {
  public Node root;

  // function to print Right view of
  // binary tree
  public  void rightView(Node root)
  {
    if (root == null) {
      return;
    }

    Queue<Node > q = new Queue<Node>();
    q.Enqueue(root);

    while (q.Count!=0) {

      // get number of nodes for each level
      int n = q.Count;

      // traverse all the nodes of the current level
      for (int i = 0; i < n; i++) {
        Node curr = q.Peek();
        q.Dequeue();

        // print the last node of each level
        if (i == n - 1) {
          Console.Write(curr.data);
          Console.Write(" ");
        }

        // if left child is not null add it into
        // the
        // queue
        if (curr.left != null) {
          q.Enqueue(curr.left);
        }

        // if right child is not null add it into
        // the
        // queue
        if (curr.right != null) {
          q.Enqueue(curr.right);
        }
      }
    }
  }

  // Driver Code
  public static void Main()
  {
    // Let's construct the tree as
    // shown in example
    BinaryTree tree = new BinaryTree();
    tree.root = new Node(1);
    tree.root.left = new Node(2);
    tree.root.right = new Node(3);
    tree.root.left.left = new Node(4);
    tree.root.left.right = new Node(5);
    tree.root.right.left = new Node(6);
    tree.root.right.right = new Node(7);
    tree.root.right.left.right = new Node(8);

    tree.rightView(tree.root);

  }
}

// This code is contributed by jana_sayantan.

Javascript

<script>

    // JavaScript program to print left view of Binary Tree

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

    // Utility function to create a new tree node
    function newNode(data)
    {
        let temp = new Node(data);
        return temp;
    }

    // function to print Right view of
    // binary tree
    function printRightView(root)
    {
        if (root == null)
            return;

        let q = [];
        q.push(root);

        while (q.length > 0) {
            // get number of nodes for each level
            let n = q.length;

            // traverse all the nodes of the current level
            while (n-- > 0) {

                let x = q[0];
                q.shift();

                // print the last node of each level
                if (n == 0) {
                    document.write(x.data + " ");
                }
                // if left child is not null push it into the
                // queue
                if (x.left != null)
                    q.push(x.left);
                // if right child is not null push it into the
                // queue
                if (x.right != null)
                    q.push(x.right);
            }
        }
    }

    // Let's construct the tree as
    // shown in example

    let root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
    root.right.left.right = newNode(8);

    printRightView(root);

</script>

Output

1 3 7 8 

Time Complexity: O(n), where n is the number of nodes in the binary tree.

Auxiliary Space: O(n) since using auxiliary space for queue

This article is contributed by Biswajit Rajak. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above