Solve sliding window and tree BFS
Company: Meta
Role: Software Engineer
Category: Coding & Algorithms
Difficulty: medium
Interview Round: Technical Screen
##### Question
Solve a typical medium-level sliding-window problem (e.g., longest substring without repeating characters). LeetCode 102. Binary Tree Level Order Traversal
https://leetcode.com/problems/binary-tree-level-order-traversal/description/
Quick Answer: This interview question evaluates algorithm design, data structures, correctness, complexity, edge cases, and implementation details in a realistic interview setting. A strong answer for Solve sliding window and tree BFS states assumptions, handles edge cases, explains trade-offs, and shows how to validate the result clearly.
Solution
# Solution Alignment
The prompt asks for an implementation-level answer. The safest way to present it is to define the state, maintain clear invariants, then walk through complexity and tests.
## Problem Restatement
##### Question Solve a typical medium-level sliding-window problem (e.g., longest substring without repeating characters). LeetCode 102. Binary Tree Level Order Traversal https://leetcode.com/problems/binary-tree-level-order-traversal/description/
## Recommended Approach
Choose traversal based on the required output. DFS is natural for subtree computations, reconstruction, and range pruning; BFS is natural for level order or side views. Keep per-depth or per-position state when the output depends on columns, rows, or depths.
## Correctness
The implementation should maintain an invariant after each loop or operation that directly matches the problem statement. At termination, that invariant implies the returned value has considered every valid candidate exactly once, or has preserved the required data-structure state after every API call.
## Complexity
Most tree traversals are O(n) time and O(h) recursion stack for DFS or O(w) queue space for BFS, where h is height and w is maximum width.
## Edge Cases and Tests
Empty tree, one node, skewed tree, duplicate values when reconstruction assumes uniqueness, deep recursion, and tie-breaking for same row/column nodes.