What is the time complexity of Binary Tree Right Side View?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Binary Tree Right Side View?

This problem uses Tree, Depth-First Search, Breadth-First Search, Binary Tree. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Binary Tree Right Side View?

Insight 1: A Breadth-First Search (BFS) or Level Order Traversal allows us to visit nodes level by level. Insight 2: The rightmost node at each level is the last node visited in that level during a BFS.

Which companies ask LeetCode 199. Binary Tree Right Side View in interviews?

This problem is frequently asked at Accolite, Flipkart, Google, J.P. Morgan, Oracle and 4 more companies.

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Binary Tree Right Side View - LeetCode 199 Solution

Q: How to solve Binary Tree Right Side View?

The solution employs a Breadth-First Search (BFS) algorithm to traverse the binary tree level by level. For each level, we iterate through all the nodes present in that level. The last node we process in each level is the rightmost node in that level, which is added to the result list. The algorithm uses a queue to maintain the order of nodes to visit during the BFS traversal.

Binary Tree Right Side View - Complete Solution Guide

Binary Tree Right Side View is LeetCode problem 199, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given the root of a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom . Example 1: Input: root = [1,2,3,null,5,null,4] Output: [1,3,4] Explanation: Example 2: Input: root = [1,2,3,4,null,null,null,5] Output: [1,3,4,5] Explanation: Example 3: Input: root = [1,null,3] Output: [1,3] Example 4: Input: root = [] Output: [] Constraints: The number of nodes in the tree is in the range [0, 100] . -100 <= Node.val <= 100

Detailed Explanation

The problem requires us to find the rightmost node at each level of a binary tree when viewed from the right side. We need to return the values of these rightmost nodes in a list, ordered from the top (root) to the bottom (leaves). The input is the root of the binary tree, and the output is a list of integers representing the right side view. The constraints specify the number of nodes is between 0 and 100, and node values are between -100 and 100.

Solution Approach

The solution employs a Breadth-First Search (BFS) algorithm to traverse the binary tree level by level. For each level, we iterate through all the nodes present in that level. The last node we process in each level is the rightmost node in that level, which is added to the result list. The algorithm uses a queue to maintain the order of nodes to visit during the BFS traversal.

Step-by-Step Algorithm

Step 1: Initialize an empty list `result` to store the right side view values.
Step 2: If the root is null (empty tree), return the empty `result` list.
Step 3: Initialize a queue and add the root node to the queue.
Step 4: While the queue is not empty, perform the following:
Step 5: Get the number of nodes in the current level using `level_size = queue.size()`.
Step 6: Iterate through the nodes in the current level using a `for` loop from `i = 0` to `level_size - 1`.
Step 7: Dequeue a node from the queue.
Step 8: If the current node is the last node in the level (i.e., `i == level_size - 1`), add its value to the `result` list.
Step 9: Enqueue the left child of the current node, if it exists.
Step 10: Enqueue the right child of the current node, if it exists.
Step 11: After the queue is empty (all nodes have been visited), return the `result` list.

Key Insights

Insight 1: A Breadth-First Search (BFS) or Level Order Traversal allows us to visit nodes level by level.
Insight 2: The rightmost node at each level is the last node visited in that level during a BFS.
Insight 3: We can utilize a queue to implement BFS efficiently.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(w)

Topics

This problem involves: Tree, Depth-First Search, Breadth-First Search, Binary Tree.

Companies

Asked at: Accolite, Flipkart, Google, J.P. Morgan, Oracle, ServiceNow, Walmart Labs, Wix, Yandex.

Binary Tree Right Side View | LeetCode 199

Medium

Tree Depth-First Search Breadth-First Search Binary Tree

Time: O(n)Space: O(w)

Solution Code

Available in:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;

class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        if (root == null) {
            return new ArrayList<>();
        }

        List<Integer> result = new ArrayList<>();
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        while (!queue.isEmpty()) {
            int levelSize = queue.size();
            for (int i = 0; i < levelSize; i++) {
                TreeNode node = queue.poll();
                if (i == levelSize - 1) {
                    result.add(node.val);
                }

                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
        }

        return result;
    }
}

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Problem Info

Description

Problem 199: Binary Tree Right Side View - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Breadth-First Search Binary Tree

Companies

AccoliteFlipkartGoogleJ.P. MorganOracleServiceNowWalmart LabsWix

Stats

Difficulty:Medium

Languages:4

Companies:9

Problem #:199

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Initialize an empty list `result` to store the right side view values.
2
Step 2: If the root is null (empty tree), return the empty `result` list.
3
Step 3: Initialize a queue and add the root node to the queue.
4
Step 4: While the queue is not empty, perform the following:
5
Step 5: Get the number of nodes in the current level using `level_size = queue.size()`.
6
Step 6: Iterate through the nodes in the current level using a `for` loop from `i = 0` to `level_size - 1`.
7
Step 7: Dequeue a node from the queue.
8
Step 8: If the current node is the last node in the level (i.e., `i == level_size - 1`), add its value to the `result` list.
9
Step 9: Enqueue the left child of the current node, if it exists.
10
Step 10: Enqueue the right child of the current node, if it exists.
11
Step 11: After the queue is empty (all nodes have been visited), return the `result` list.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(w)

📝

Why?

The time complexity is O(n) as each node is visited exactly once. The space complexity is dominated by the queue which holds at most all nodes at one level of the tree. The maximum number of nodes at any one level is proportional to the tree's width w. In a balanced tree, width can be n/2.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Depth-First Search (DFS)

A DFS approach can also solve this problem. We can perform a recursive traversal, maintaining the current level and a list of rightmost nodes seen so far. If the current level is not yet in the list, we add the node's value to the list. Then, we recursively call the function on the right subtree before the left subtree. This ensures that if a node exists on the right side, it will be added before any node on the left at the same level.

Time: O(n)Space: O(h), where h is the height of the tree. In the worst case (skewed tree), h can be n, resulting in O(n) space complexity. The space is used by the call stack during the recursive calls.

Trade-offs: DFS can be slightly more memory-efficient than BFS for balanced trees (O(log n) space), but BFS is often preferred for its iterative nature and predictable space usage based on the width of the tree, not the height. DFS can be less intuitive to implement for this specific problem.

✨

Key Insights

1Insight 1: A Breadth-First Search (BFS) or Level Order Traversal allows us to visit nodes level by level.
2Insight 2: The rightmost node at each level is the last node visited in that level during a BFS.
3Insight 3: We can utilize a queue to implement BFS efficiently.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty Tree (root is null). The solution correctly returns an empty list in this case.
Edge case 2: Skewed Tree (only left or right children). The solution correctly identifies the right side view nodes even in skewed trees because it processes each level and picks the last node.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the edge case of an empty tree (root being null).

✗Common mistake 2: Incorrectly determining the rightmost node at each level. Some might mistakenly take the right child of the root as the rightmost node of all levels. It's crucial to perform a level-order traversal to identify the rightmost node at *each* level.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Displaying a simplified outline or summary of a hierarchical structure, such as an organizational chart or a file system, by only showing the rightmost element at each level.

🏦Real-world scenario 2: In game development, visualizing the 'silhouette' of a 3D object from a particular viewpoint (in this case, the right side), which could be useful for collision detection or level design.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Most tree problems can be solved recursively. The base case is usually a null node. Think about what information needs to flow up vs down the tree.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Not handling null nodes properly. Always check if the current node exists before accessing its children.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 20, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

What is the difference between DFS and BFS for trees?

DFS (using recursion or stack) explores depth-first, useful for path-finding and backtracking problems. BFS (using queue) explores level-by-level, ideal for finding shortest paths or processing nodes by depth. Choose based on whether level information matters.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Binary Tree Right Side View"?

For this medium-level Tree problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether depth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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