Print Right View of a Binary Tree
Last Updated :
24 May, 2024
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 <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
// 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) {
console.log(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();
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
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.
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
Python
# 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
// 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) {
console.log(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);
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:
- Initialize an empty vector res to hold the result.
- Initialize a pointer curr to the root of the binary tree.
- 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. - 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
Java
import java.util.ArrayList;
import java.util.List;
// Definition for a binary tree node
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int val)
{
this.val = val;
this.left = null;
this.right = null;
}
}
public class RightSideView {
public List<Integer> rightSideView(TreeNode root)
{
List<Integer> res = new ArrayList<>();
TreeNode curr = root;
int n = 4; // You can set 'n' to any desired value
while (curr != null) {
if (curr.right
== null) { // if there is no right child,
// add the current node's value
// to the list
res.add(curr.val);
curr = curr.left; // move to the left child
}
else { // if there is a right child
TreeNode next
= curr.right; // set the next node to
// the right child
while (
next.left != null
&& next.left
!= curr) { // traverse the left
// subtree of the
// right child until a
// leaf node or the
// current node is
// reached
next = next.left;
}
if (next.left
== null) { // if the left child of the
// next node is null
res.add(curr.val);
next.left = curr;
curr = curr.right;
}
else {
next.left = null;
curr = curr.left;
}
}
}
return res.subList(0, Math.min(n, res.size()));
}
// Driver Code
public static void main(String[] args)
{
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);
RightSideView rightSideView = new RightSideView();
List<Integer> res
= rightSideView.rightSideView(root);
System.out.println(res);
}
}
// This code is contributed by Veerendra_Singh_Rajpoot
Python
# 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
C#
using System;
using System.Collections.Generic;
// Definition for a binary tree node
public class TreeNode
{
public int val;
public TreeNode left;
public TreeNode right;
public TreeNode(int x)
{
val = x;
left = null;
right = null;
}
}
public class RightSideView
{
public List<int> MorrisTraversal(TreeNode root)
{
List<int> res = new List<int>();
TreeNode curr = root;
while (curr != null)
{
if (curr.right == null)
{
// If there is no right child, add the current node's value to the list
res.Add(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
// Traverse the left subtree of the right child until a leaf node or the current node is reached
while (next.left != null && next.left != curr)
{
next = next.left;
}
if (next.left == null)
{
// If the left child of the next node is null
res.Add(curr.val);
next.left = curr;
curr = curr.right;
}
else
{
next.left = null;
curr = curr.left;
}
}
}
return res;
}
// Driver code
public static void Main(string[] args)
{
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);
RightSideView rightSideView = new RightSideView();
List<int> res = rightSideView.MorrisTraversal(root);
foreach (int val in res)
{
Console.Write(val + " ");
}
Console.WriteLine();
}
}
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.
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.
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