Print Right View of a Binary Tree

Given a Binary Tree, print the Right view of it. 

The right view of a Binary Tree is a set of nodes visible when the tree is visited from the Right side.

Examples:

Input:
          1
       /     \
     2        3
   /   \       /  \
  4     5   6    7
                 \
                   8
Output: Right view of the tree is 1 3 7 8

Input:
          1
       /     
     8        
   / 
  7
Output: Right view of the tree is 1 8 7

Right View of a Binary Tree using Recursion:

The idea is to use recursion and keep track of the maximum level also. And traverse the tree in a manner that the right subtree is visited before the left subtree.

Follow the steps below to solve the problem:

• Perform Postorder traversal to get the rightmost node first
• Maintain a variable name max_level which will store till which it prints the right view
• While traversing the tree in a postorder manner if the current level is greater than max_level then print the current node and update max_level by the current level

Below is the implementation of the above 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: O(N), Traversing the Tree having N nodes
Auxiliary Space: O(N), Function Call stack space in the worst case.

Right View of a Binary Tree using Level Order Traversal:

The idea is to use Level Order Traversal as the last node every level gives the right view of the binary tree.

Follow the steps below to solve the problem:

• Perform level order traversal on the tree
• At every level print the last node of that 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

Right View of a Binary Tree using Morris Traversal:

Follow the Steps to implement the approach:

1. Initialize an empty vector res to hold the result.
2. Initialize a pointer curr to the root of the binary tree.
3. While curr is not null, do the following:
   If curr has no right child:
   Add the value of curr to res.
   Set curr to its right child.
   Otherwise:
   Find the inorder successor of curr by initializing a pointer next to the right child of curr, and then repeatedly moving left until next has no left child or its left child is equal to curr.
   If next has no left child:
   Add the value of curr to res.
   Set the left child of next to curr.
   Set curr to its right child.
   Otherwise:
   Set the left child of next to null.
   Set curr to its left child.
4. Return res, which contains the last node at each level of the binary tree as seen from the right side.

Below is the implementation of the above approach.

C++

// C++ code to implement the morris traversal approach
#include <iostream>
#include <vector>

using namespace std;

// Definition for a binary tree node
struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x)
        : val(x)
        , left(NULL)
        , right(NULL)
    {
    }
};

vector<int> rightSideView(TreeNode* root)
{
    vector<int> res;
    TreeNode* curr = root;
    while (curr) {
        if (!curr->right) { // if there is no right child ,
                            // add the current node's value
                            // to the vector
            res.push_back(curr->val);
            curr = curr->right; // move to the right child
        }
        else { // if there is a right child
            TreeNode* next
                = curr->right; // set the next node to the
                               // right child
            while (next->left
                   && next->left
                          != curr) { // traverse the left
                                     // subtree of the right
                                     // child untill a leaf
                                     // node or the current
                                     // node is reached
                next = next->left;
            }

            if (!next->left) { // if the left child of the
                               // next node is NULL
                res.push_back(curr->val);
                next->left = curr;
                curr = curr->right;
            }
            else {
                next->left = NULL;
                curr = curr->left;
            }
        }
    }
    return res;
}
// Driver code
int main()
{
    TreeNode* root = new TreeNode(1);
    root->left = new TreeNode(2);
    root->right = new TreeNode(3);
    root->left->left = new TreeNode(4);
    root->left->right = new TreeNode(5);
    root->right->left = new TreeNode(6);
    root->right->right = new TreeNode(7);
    root->right->right->right = new TreeNode(8);
    vector<int> res = rightSideView(root);
    for (int i : res) {
        cout << i << " ";
    }
    cout << endl;

    return 0;
}
// This code is contributed by Veerendra_Singh_Rajpoot

Python3

# Python code to implement the morris traversal approach

# Definition for a binary tree node


class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None


def rightSideView(root):
    res = []
    curr = root
    while curr:
        if not curr.right:  # if there is no right child,
                            # add the current node's value
                            # to the vector
            res.append(curr.val)
            curr = curr.right  # move to the right child
        else:  # if there is a right child
            next = curr.right  # set the next node to the
            # right child
            while next.left and next.left != curr:
                # traverse the left subtree of the right
                # child untill a leaf node or the current
                # node is reached
                next = next.left

            if not next.left:  # if the left child of the
                              # next node is NULL
                res.append(curr.val)
                next.left = curr
                curr = curr.right
            else:
                next.left = None
                curr = curr.left
    return res


# Driver code
if __name__ == '__main__':
    root = TreeNode(1)
    root.left = TreeNode(2)
    root.right = TreeNode(3)
    root.left.left = TreeNode(4)
    root.left.right = TreeNode(5)
    root.right.left = TreeNode(6)
    root.right.right = TreeNode(7)
    root.right.right.right = TreeNode(8)
    res = rightSideView(root)
    for i in res:
        print(i, end=" ")
    print()
# Contributed by adityasha4x71

Javascript

// Javascript code addition

// Definition for a binary tree node
class TreeNode {
    constructor(val) {
        this.val = val;
        this.left = null;
        this.right = null;
    }
}

function rightSideView(root) {
    const res = [];
    let curr = root;
    let n = 4;
    while (curr) {
        if (!curr.right) { // if there is no right child ,
                            // add the current node's value
                            // to the array
            res.push(curr.val);
            curr = curr.left; // move to the left child
        }
        else { // if there is a right child
            let next = curr.right; // set the next node to the
                                    // right child
            while (next.left && next.left !== curr) { // traverse the left
                                                      // subtree of the right
                                                      // child untill a leaf
                                                      // node or the current
                                                      // node is reached
                next = next.left;
            }

            if (!next.left) { // if the left child of the
                               // next node is NULL
                res.push(curr.val);
                next.left = curr;
                curr = curr.right;
            }
            else {
                next.left = null;
                curr = curr.left;
            }
        }
    }
    return res.slice(0, n);
}

// Driver code
const root = new TreeNode(1);
root.left = new TreeNode(2);
root.right = new TreeNode(3);
root.left.left = new TreeNode(4);
root.left.right = new TreeNode(5);
root.right.left = new TreeNode(6);
root.right.right = new TreeNode(7);
root.right.right.right = new TreeNode(8);

const res = rightSideView(root);
console.log(res);

// The code is contributed by Arushi Goel.

Output

1 3 7 8 

Time Complexity: O(n) , The time complexity of the Morris Traversal approach is O(n), where n is the number of nodes in the binary tree. This is because we visit each node exactly twice (once when we find its inorder predecessor, and once when we visit it from its inorder predecessor).

Auxiliary Space:  O(1) , The space complexity of the Morris Traversal approach is O(1), because we only use a constant amount of extra space for the res, curr, and next pointers. We do not use any additional data structures or recursive function calls that would increase the space complexity.

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