What is the time complexity of Find Largest Value in Each Tree Row?

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

What data structures are used to solve Find Largest Value in Each Tree Row?

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 Find Largest Value in Each Tree Row?

Insight 1: Level-order traversal (Breadth-First Search - BFS) is the most suitable approach to process the tree level by level. Insight 2: Using a queue data structure is essential for implementing level-order traversal efficiently.

Which companies ask LeetCode 515. Find Largest Value in Each Tree Row in interviews?

This problem is frequently asked at LinkedIn.

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Find Largest Value in Each Tree Row - LeetCode 515 Solution

Q: How to solve Find Largest Value in Each Tree Row?

The provided solutions use Breadth-First Search (BFS) to traverse the binary tree level by level. A queue is used to store the nodes at each level. For each level, the algorithm iterates through the nodes in the queue (representing that level), finds the maximum value among those nodes, and adds that maximum value to the result list. Then, it enqueues the children of each node (left and right) to prepare for the next level's processing. This continues until the queue is empty, indicating that all nodes in the tree have been visited.

Find Largest Value in Each Tree Row - Complete Solution Guide

Find Largest Value in Each Tree Row is LeetCode problem 515, 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, return an array of the largest value in each row of the tree (0-indexed) . Example 1: Input: root = [1,3,2,5,3,null,9] Output: [1,3,9] Example 2: Input: root = [1,2,3] Output: [1,3] Constraints: The number of nodes in the tree will be in the range [0, 10 4 ] . -2 31 <= Node.val <= 2 31 - 1

Detailed Explanation

The problem requires you to traverse a binary tree and find the largest value at each level (or row) of the tree. The input is the root node of a binary tree. The output is a list (or array) where each element represents the largest value found in the corresponding level of the tree. The levels are numbered starting from 0 at the root. The constraints specify the size of the tree (number of nodes) and the range of values for each node.

Solution Approach

The provided solutions use Breadth-First Search (BFS) to traverse the binary tree level by level. A queue is used to store the nodes at each level. For each level, the algorithm iterates through the nodes in the queue (representing that level), finds the maximum value among those nodes, and adds that maximum value to the result list. Then, it enqueues the children of each node (left and right) to prepare for the next level's processing. This continues until the queue is empty, indicating that all nodes in the tree have been visited.

Step-by-Step Algorithm

Step 1: Initialize an empty list called `result` to store the largest values for each level.
Step 2: Check if the root is null. If it is, 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, repeat steps 5-9.
Step 5: Get the number of nodes in the current level by checking the queue's size.
Step 6: Initialize a variable `max_in_level` to the smallest possible integer value (negative infinity in Python, Integer.MIN_VALUE in Java/C++, INT_MIN in C).
Step 7: Iterate through the nodes in the current level (from 0 to level_size - 1).
Step 8: Dequeue a node from the queue. Update `max_in_level` to be the maximum of `max_in_level` and the current node's value.
Step 9: Enqueue the left and right children of the current node (if they exist).
Step 10: After processing all nodes in the current level, add `max_in_level` to the `result` list.
Step 11: Once the queue is empty, return the `result` list.

Key Insights

Insight 1: Level-order traversal (Breadth-First Search - BFS) is the most suitable approach to process the tree level by level.
Insight 2: Using a queue data structure is essential for implementing level-order traversal efficiently.
Insight 3: It's important to keep track of the maximum value for each level as you traverse the nodes within that level.

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: LinkedIn.

Find Largest Value in Each Tree Row | LeetCode 515

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.*;

class Solution {
    public List<Integer> largestValues(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();
            int maxInLevel = Integer.MIN_VALUE;

            for (int i = 0; i < levelSize; i++) {
                TreeNode node = queue.poll();
                maxInLevel = Math.max(maxInLevel, node.val);

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

            result.add(maxInLevel);
        }

        return result;
    }
}

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Problem Info

Description

Problem 515: Find Largest Value in Each Tree Row - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Breadth-First Search Binary Tree

Companies

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:515

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 called `result` to store the largest values for each level.
2
Step 2: Check if the root is null. If it is, 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, repeat steps 5-9.
5
Step 5: Get the number of nodes in the current level by checking the queue's size.
6
Step 6: Initialize a variable `max_in_level` to the smallest possible integer value (negative infinity in Python, Integer.MIN_VALUE in Java/C++, INT_MIN in C).
7
Step 7: Iterate through the nodes in the current level (from 0 to level_size - 1).
8
Step 8: Dequeue a node from the queue. Update `max_in_level` to be the maximum of `max_in_level` and the current node's value.
9
Step 9: Enqueue the left and right children of the current node (if they exist).
10
Step 10: After processing all nodes in the current level, add `max_in_level` to the `result` list.
11
Step 11: Once the queue is empty, 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) because each node is enqueued and dequeued exactly once. The space complexity is O(w) because the queue's maximum size is determined by the width of the tree at its widest level. If the tree is very unbalanced, the space complexity will be closer to O(1). If the tree is complete or nearly complete, the space complexity will approach O(n).

📖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)

It's possible to solve this problem using Depth-First Search (DFS) with level information. A recursive function can be used to traverse the tree. A list is maintained to store the largest value at each level. The recursive function takes the current node, the current level, and the list as arguments. If the current level is beyond the size of the list, append the node's value to the list. Otherwise, update the list's value at that level with the maximum of the existing value and the node's value.

Time: O(n), where n is the number of nodes in the tree.Space: O(h), where h is the height of the tree. In the worst case (skewed tree), h can be n, making the space complexity O(n).

Trade-offs: DFS might be slightly more space-efficient than BFS for balanced trees (O(h) vs O(w)), but it can be less efficient for skewed trees due to recursion depth. BFS is generally preferred for level-order traversal problems.

✨

Key Insights

1Insight 1: Level-order traversal (Breadth-First Search - BFS) is the most suitable approach to process the tree level by level.
2Insight 2: Using a queue data structure is essential for implementing level-order traversal efficiently.
3Insight 3: It's important to keep track of the maximum value for each level as you traverse the nodes within that level.

📚

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 handles this case by returning an empty list.
Edge case 2: Tree with only one node (the root). The solution correctly adds the root's value to the result and returns a list containing only that value.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not initializing `max_in_level` to the smallest possible integer value (e.g., 0). This can lead to incorrect results if all node values at a level are negative.

✗Common mistake 2: Not handling the case where the root is null, which can lead to a null pointer exception or incorrect output.

✅

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: Analyzing organizational charts to find the highest-paid employee at each level of management.

🏦Real-world scenario 2: Processing game tree structures to determine the best possible move at each turn, maximizing the player's score.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: November 3, 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

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

@DevExpert92Jul 14

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

@AlgoMasterJul 7

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 30

❓

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 "Find Largest Value in Each Tree Row"?

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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