Skip to content
Related Articles

Related Articles

Improve Article
Print Right View of a Binary Tree
  • Difficulty Level : Medium
  • Last Updated : 22 Jun, 2021

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->left->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

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).



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
            // stack
            if (x->left)
                q.push(x->left);
            // if right child is not null push it into the
            // stack
            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

Output:

1 3 7 8

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

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer DSA Live Classes 




My Personal Notes arrow_drop_up
Recommended Articles
Page :