What is the time complexity of Binary Tree Level Order Traversal II?

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

What data structures are used to solve Binary Tree Level Order Traversal II?

This problem uses Tree, 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 Level Order Traversal II?

Insight 1: Level order traversal is best implemented using Breadth-First Search (BFS) with a queue. Insight 2: To achieve bottom-up order, we can perform a regular level order traversal and then reverse the resulting list of levels.

Which companies ask LeetCode 107. Binary Tree Level Order Traversal II in interviews?

This problem is frequently asked at Google, Meta, Revolut.

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Binary Tree Level Order Traversal II - LeetCode 107 Solution

Q: How to solve Binary Tree Level Order Traversal II?

The solution uses a standard Breadth-First Search (BFS) algorithm to traverse the binary tree level by level. A queue is used to store the nodes at each level. For each level, we iterate through all nodes in the current level of the queue, add their values to a list representing that level, and enqueue their children (if they exist). After processing all levels, the list of levels is reversed to achieve the desired bottom-up order.

Binary Tree Level Order Traversal II - Complete Solution Guide

Binary Tree Level Order Traversal II is LeetCode problem 107, 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 the bottom-up level order traversal of its nodes' values . (i.e., from left to right, level by level from leaf to root). Example 1: Input: root = [3,9,20,null,null,15,7] Output: [[15,7],[9,20],[3]] Example 2: Input: root = [1] Output: [[1]] Example 3: Input: root = [] Output: [] Constraints: The number of nodes in the tree is in the range [0, 2000] . -1000 <= Node.val <= 1000

Detailed Explanation

The problem asks us to perform a level order traversal of a binary tree, but instead of returning the levels from the root to the leaves, we need to return them in reverse order, i.e., from the leaves to the root. The input is the root node of a binary tree, and the output is a list of lists, where each inner list represents a level in the tree, starting from the bottom (leaves) and going up to the root.

Solution Approach

The solution uses a standard Breadth-First Search (BFS) algorithm to traverse the binary tree level by level. A queue is used to store the nodes at each level. For each level, we iterate through all nodes in the current level of the queue, add their values to a list representing that level, and enqueue their children (if they exist). After processing all levels, the list of levels is reversed to achieve the desired bottom-up order.

Step-by-Step Algorithm

Step 1: Handle the base case: If the root is null, return an empty list.
Step 2: Initialize an empty list `result` to store the levels and a queue `queue` with the root node.
Step 3: While the queue is not empty, repeat the following steps:
Step 4: Determine the number of nodes in the current level using `level_size = len(queue)`.
Step 5: Initialize an empty list `current_level` to store the values of the nodes in the current level.
Step 6: Iterate `level_size` times:
Step 7: Dequeue a node from the queue.
Step 8: Append the node's value to `current_level`.
Step 9: Enqueue the node's left child (if it exists).
Step 10: Enqueue the node's right child (if it exists).
Step 11: Append `current_level` to `result`.
Step 12: Reverse the `result` list to get the bottom-up level order traversal and return it.

Key Insights

Insight 1: Level order traversal is best implemented using Breadth-First Search (BFS) with a queue.
Insight 2: To achieve bottom-up order, we can perform a regular level order traversal and then reverse the resulting list of levels.
Insight 3: Efficient queue management is crucial for good performance, especially with large trees.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

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

Companies

Asked at: Google, Meta, Revolut.

Binary Tree Level Order Traversal II | LeetCode 107

Medium

Tree Breadth-First Search Binary Tree

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

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<List<Integer>> levelOrderBottom(TreeNode root) {
        if (root == null) {
            return new ArrayList<>();
        }

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

        while (!queue.isEmpty()) {
            int levelSize = queue.size();
            List<Integer> currentLevel = new ArrayList<>();
            for (int i = 0; i < levelSize; i++) {
                TreeNode node = queue.poll();
                currentLevel.add(node.val);
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
            result.add(currentLevel);
        }

        List<List<Integer>> reversedResult = new ArrayList<>();
        for (int i = result.size() - 1; i >= 0; i--) {
            reversedResult.add(result.get(i));
        }
        return reversedResult;
    }
}

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Problem Info

Description

Problem 107: Binary Tree Level Order Traversal II - Detailed explanation and solution approaches.

Topics

Tree Breadth-First Search Binary Tree

Companies

GoogleMetaRevolut

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:107

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

12 steps

1
Step 1: Handle the base case: If the root is null, return an empty list.
2
Step 2: Initialize an empty list `result` to store the levels and a queue `queue` with the root node.
3
Step 3: While the queue is not empty, repeat the following steps:
4
Step 4: Determine the number of nodes in the current level using `level_size = len(queue)`.
5
Step 5: Initialize an empty list `current_level` to store the values of the nodes in the current level.
6
Step 6: Iterate `level_size` times:
7
Step 7: Dequeue a node from the queue.
8
Step 8: Append the node's value to `current_level`.
9
Step 9: Enqueue the node's left child (if it exists).
10
Step 10: Enqueue the node's right child (if it exists).
11
Step 11: Append `current_level` to `result`.
12
Step 12: Reverse the `result` list to get the bottom-up level order traversal and return it.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because we visit each node once during the BFS traversal. The space complexity is O(n) because, in the worst-case scenario, the queue will store approximately half the nodes of the tree at its widest level (for a complete binary tree). Additionally, the 'result' list stores all the node values.

📖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

Recursive Depth-First Search (DFS)

An alternative approach is to use a recursive DFS. Maintain a list of lists representing the levels. In the recursive function, pass the current node and the level. If the level doesn't exist in the list of lists, create it. Append the node's value to the appropriate level list. Recursively call the function for the left and right children, incrementing the level. Finally reverse the list of lists.

Time: O(n)Space: O(n) in the worst case (skewed tree), where n is the height of the tree. It also uses O(n) for the result list.

Trade-offs: DFS can be more complex to implement for level order traversal than BFS, especially for bottom-up traversal. BFS is generally preferred due to its iterative nature and easier management of levels. Recursive solutions might hit recursion limit on very deep trees.

✨

Key Insights

1Insight 1: Level order traversal is best implemented using Breadth-First Search (BFS) with a queue.
2Insight 2: To achieve bottom-up order, we can perform a regular level order traversal and then reverse the resulting list of levels.
3Insight 3: Efficient queue management is crucial for good performance, especially with large trees.

📚

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 by returning an empty list.
Edge case 2: Tree with only one node. The solution handles this correctly by returning a list containing a single list with the root's value.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the empty tree case (root is null).

✗Common mistake 2: Not correctly tracking the level size in the BFS loop, which can lead to incorrect level separation.

✗Common mistake 3: Failing to reverse the list of levels after the BFS traversal to achieve the bottom-up order.

✅

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 hierarchical data in reverse order, such as organizational charts or file system structures, starting from the leaves.

🏦Real-world scenario 2: Processing events or data in a tree-like structure where dependencies are resolved from the bottom up, such as in dependency injection frameworks or build systems.

💡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.

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: September 16, 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

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

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

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

@TechLeadJun 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 Level Order Traversal II"?

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 breadth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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