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

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

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

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 Zigzag Level Order Traversal?

Insight 1: Level order traversal is best achieved using Breadth-First Search (BFS). Insight 2: A queue is suitable for managing the nodes at each level in BFS.

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Binary Tree Zigzag Level Order Traversal - LeetCode 103 Solution

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

The provided code implements a Breadth-First Search (BFS) algorithm using a queue to traverse the binary tree level by level. It maintains a boolean variable `left_to_right` to keep track of the current traversal direction. For each level, it iterates through the nodes in the queue, adding their values to a temporary deque. If `left_to_right` is true, it adds the value to the end of the deque; otherwise, it adds the value to the beginning. After processing all nodes at a level, it converts the deque to a list and adds it to the result list. Finally, it flips the `left_to_right` flag to switch the traversal direction for the next level.

Q: Which companies ask LeetCode 103. Binary Tree Zigzag Level Order Traversal in interviews?

This problem is frequently asked at Adobe, Bloomberg, ByteDance, Citadel, Flipkart and 12 more companies.

Binary Tree Zigzag Level Order Traversal - Complete Solution Guide

Binary Tree Zigzag Level Order Traversal is LeetCode problem 103, 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 zigzag level order traversal of its nodes' values . (i.e., from left to right, then right to left for the next level and alternate between). Example 1: Input: root = [3,9,20,null,null,15,7] Output: [[3],[20,9],[15,7]] 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] . -100 <= Node.val <= 100

Detailed Explanation

The problem asks us to perform a level order traversal of a binary tree, but with a twist. Instead of always traversing each level from left to right, we need to alternate the traversal direction. The first level is traversed from left to right, the second from right to left, the third from left to right, and so on, creating a 'zigzag' pattern. 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 of the tree, with the node values arranged according to the zigzag pattern.

Solution Approach

The provided code implements a Breadth-First Search (BFS) algorithm using a queue to traverse the binary tree level by level. It maintains a boolean variable `left_to_right` to keep track of the current traversal direction. For each level, it iterates through the nodes in the queue, adding their values to a temporary deque. If `left_to_right` is true, it adds the value to the end of the deque; otherwise, it adds the value to the beginning. After processing all nodes at a level, it converts the deque to a list and adds it to the result list. Finally, it flips the `left_to_right` flag to switch the traversal direction for the next level.

Step-by-Step Algorithm

Step 1: Initialize an empty list called `result` to store the final zigzag level order traversal.
Step 2: Handle the base case: if the root is null, return the empty `result` list.
Step 3: Initialize a queue with the root node.
Step 4: Initialize a boolean variable `left_to_right` to true, indicating the initial traversal direction.
Step 5: While the queue is not empty:
Step 6: Get the number of nodes at the current level (level size).
Step 7: Initialize an empty deque `current_level` to store the node values for the current level.
Step 8: Iterate `level_size` times:
Step 9: Dequeue a node from the queue.
Step 10: If `left_to_right` is true, append the node's value to the end of `current_level`. Otherwise, append the node's value to the beginning of `current_level`.
Step 11: Enqueue the node's left and right children (if they exist).
Step 12: Convert the deque `current_level` to a list and append it to the `result` list.
Step 13: Flip the value of `left_to_right` to switch the traversal direction for the next level.
Step 14: Return the `result` list.

Key Insights

Insight 1: Level order traversal is best achieved using Breadth-First Search (BFS).
Insight 2: A queue is suitable for managing the nodes at each level in BFS.
Insight 3: Using a Deque (Double-Ended Queue) allows efficient insertion and removal from both ends, making it ideal for handling the alternating direction of traversal.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

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

Companies

Asked at: Adobe, Bloomberg, ByteDance, Citadel, Flipkart, Intuit, LinkedIn, Meta, Microsoft, Nutanix, Oracle, ServiceNow, Sigmoid, TikTok, Walmart Labs, Yandex, eBay.

Binary Tree Zigzag Level Order Traversal | LeetCode 103

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.List;
import java.util.ArrayList;
import java.util.Deque;
import java.util.LinkedList;

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

        List<List<Integer>> result = new ArrayList<>();
        Deque<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        boolean leftToRight = true;

        while (!queue.isEmpty()) {
            int levelSize = queue.size();
            Deque<Integer> currentLevel = new LinkedList<>();

            for (int i = 0; i < levelSize; i++) {
                TreeNode node = queue.poll();

                if (leftToRight) {
                    currentLevel.offer(node.val);
                } else {
                    currentLevel.offerFirst(node.val);
                }

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

            result.add(new ArrayList<>(currentLevel));
            leftToRight = !leftToRight;
        }

        return result;
    }
}

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Problem Info

Description

Problem 103: Binary Tree Zigzag Level Order Traversal - Detailed explanation and solution approaches.

Topics

Tree Breadth-First Search Binary Tree

Companies

AdobeBloombergByteDanceCitadelFlipkartIntuitLinkedInMeta

Stats

Difficulty:Medium

Languages:4

Companies:17

Problem #:103

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

14 steps

1
Step 1: Initialize an empty list called `result` to store the final zigzag level order traversal.
2
Step 2: Handle the base case: if the root is null, return the empty `result` list.
3
Step 3: Initialize a queue with the root node.
4
Step 4: Initialize a boolean variable `left_to_right` to true, indicating the initial traversal direction.
5
Step 5: While the queue is not empty:
6
Step 6: Get the number of nodes at the current level (level size).
7
Step 7: Initialize an empty deque `current_level` to store the node values for the current level.
8
Step 8: Iterate `level_size` times:
9
Step 9: Dequeue a node from the queue.
10
Step 10: If `left_to_right` is true, append the node's value to the end of `current_level`. Otherwise, append the node's value to the beginning of `current_level`.
11
Step 11: Enqueue the node's left and right children (if they exist).
12
Step 12: Convert the deque `current_level` to a list and append it to the `result` list.
13
Step 13: Flip the value of `left_to_right` to switch the traversal direction for the next level.
14
Step 14: 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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because each node is visited and processed exactly once. The space complexity is O(n) because in the worst-case scenario (a complete binary tree), the queue might contain approximately n/2 nodes, which is still 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, though less intuitive, to implement this using DFS. You would keep track of the level and append to the result list at the correct index. Then, based on the level and direction you can either append to the end or insert at the beginning of the sublist

Time: O(n)Space: O(n) in the worst case (skewed tree)

Trade-offs: BFS is typically preferred because it is more straightforward for level-order traversal. DFS requires additional logic to handle the level information and direction.

✨

Key Insights

1Insight 1: Level order traversal is best achieved using Breadth-First Search (BFS).
2Insight 2: A queue is suitable for managing the nodes at each level in BFS.
3Insight 3: Using a Deque (Double-Ended Queue) allows efficient insertion and removal from both ends, making it ideal for handling the alternating direction of traversal.

📚

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 code correctly handles this case by returning an empty list.
Edge case 2: Single-node tree. The code correctly returns a list containing a list with the single node's value.
Edge case 3: Skewed tree (only left or right children). The code correctly traverses and zigzags through the skewed tree.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the edge case of an empty tree.

✗Common mistake 2: Not correctly switching the direction of traversal after each level, leading to all levels being traversed in the same direction.

✗Common mistake 3: Using the wrong data structure for `current_level`. A `List` would require `insert(0, value)` for right-to-left traversal, which is O(n), increasing the overall time complexity. A `Deque` is the right choice as both `append` and `appendleft` are O(1).

✅

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: UI frameworks often use tree structures to represent the component hierarchy. A zigzag traversal could be used to apply alternating styles or effects to elements at different levels of the UI.

🏦Real-world scenario 2: Database indexing: Tree-based indexes can be visualized using zigzag traversals for debugging or analysis purposes.

💡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: September 12, 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

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

@CodeNinjaJul 13

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

@TechLeadJul 6

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

@DevExpert92Jun 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 Zigzag Level Order Traversal"?

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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