The solution to the Binary Tree Level Order Traversal problem relies on performing a breadth-first search (BFS) using a queue. The process involves iterating through each level and gathering the values in order, starting from the root and moving down level by level.
Problem Statement
Given the root of a binary tree, the task is to return the level order traversal of its nodes' values. This means you should traverse the tree from left to right, one level at a time, collecting all node values at each level.
For example, given a tree with root node value 3 and its children, the correct output should represent the nodes grouped by their respective levels. This problem requires efficient traversal using an appropriate approach like BFS, which can be achieved using a queue.
Examples
Example 1
Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[15,7]]
Example details omitted.
Example 2
Input: root = [1]
Output: [[1]]
Example details omitted.
Example 3
Input: root = []
Output: []
Example details omitted.
Constraints
- The number of nodes in the tree is in the range [0, 2000].
- -1000 <= Node.val <= 1000
Solution Approach
BFS with Queue
The most intuitive approach for this problem is using breadth-first search (BFS) with a queue. Start by enqueueing the root node. Then, for each level, dequeue nodes one by one, record their values, and enqueue their children (if any). This process ensures that the traversal happens level by level, left to right.
Queue Operations for Level Tracking
Using the queue to track nodes is crucial for managing the traversal levels. At each iteration, check the queue's size to determine how many nodes are at the current level. Process these nodes, then push their children into the queue for subsequent levels. This guarantees the BFS flow and keeps nodes grouped by their respective levels.
Edge Cases and Constraints
Handle edge cases where the tree is empty (i.e., no nodes). In such cases, return an empty list. Also, consider trees with only one node or a deep unbalanced tree. Since the number of nodes can be up to 2000, ensure that the algorithm performs efficiently within the problem's constraints.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time complexity is O(n), where n is the number of nodes in the binary tree, since we visit each node once. The space complexity is O(n) due to the queue used to store nodes level by level, which can grow up to the number of nodes at the widest level in the tree.
What Interviewers Usually Probe
- Do you understand how to perform BFS traversal using a queue?
- Can you explain why BFS is appropriate for this problem?
- Are you able to handle edge cases like an empty tree or a tree with a single node?
Common Pitfalls or Variants
Common pitfalls
- Misunderstanding the level order traversal concept and treating it as a depth-first traversal.
- Failing to properly manage the queue to ensure that nodes are processed level by level.
- Not accounting for edge cases such as an empty tree or single-node tree, which require special handling.
Follow-up variants
- Spiral Level Order Traversal
- Zigzag Level Order Traversal
- Reverse Level Order Traversal
How GhostInterview Helps
- Screenshot or capture input to identify the binary tree structure for BFS traversal.
- Answer path with BFS queue management and space-time complexity explained clearly.
- Supports screen-share workflows for step-by-step BFS traversal demonstration.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the primary approach for solving Binary Tree Level Order Traversal?
The problem can be solved using breadth-first search (BFS) with a queue. This allows you to traverse nodes level by level.
How do we handle edge cases in Binary Tree Level Order Traversal?
Edge cases like an empty tree should return an empty list, while a tree with only one node should return that node in a list.
What is the time complexity of the Binary Tree Level Order Traversal solution?
The time complexity is O(n), where n is the number of nodes in the tree, as we process each node once during BFS.
Can you explain the space complexity for Binary Tree Level Order Traversal?
The space complexity is O(n), where n is the number of nodes in the tree, as we store nodes in the queue at each level.
What happens if we do not use BFS for Binary Tree Level Order Traversal?
Without BFS, we risk processing nodes in an incorrect order. BFS ensures nodes are processed level by level, maintaining the required order.
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