Databricks·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

May 2026

Summary

Databricks coding round, one problem the whole time. File system tree traversal, nothing too wild, but the setup took a minute to wrap my head around.

Questions Asked (1)

You're given a file system modeled as a tree. Directory nodes can have children (other directories or files), and file nodes have an is_encrypted flag. Write a function that recursively traverses the tree from a given directory and returns a tuple of (encrypted_count, unencrypted_count).

Algorithms & Data StructuresSystem Design

Author's notes

Pretty clean once I stopped second-guessing the base case.

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AI HintsAI Generated

Suggested Approach

Clarify the tree node structure and then outline a recursive post-order traversal that accumulates counts from children. For each node, if it's a file, increment the appropriate counter based on is_encrypted; if it's a directory, sum the counts from recursive calls. Finally, return the aggregated tuple.

Pro tip: Mention that you'd use an iterative approach with an explicit stack if recursion depth is a concern, and discuss how to handle edge cases like empty directories or permission errors.

1. Clarify the data model

Ask about the node structure: does each node have a type (file/directory), children list, and is_encrypted flag? Confirm that only files have is_encrypted.

2. Define the recursive function

Write a function that takes a node and returns (encrypted_count, unencrypted_count). For a file, return (1,0) or (0,1) based on is_encrypted; for a directory, initialize counts to (0,0).

3. Traverse children and aggregate

For each child of a directory, recursively call the function and add the returned counts to the directory's counts. This is a post-order traversal.

4. Handle edge cases

Consider empty directories, null children, and potential cycles (if the tree is not strictly a tree). Discuss error handling for inaccessible nodes.

5. Analyze complexity and alternatives

State that time complexity is O(n) where n is number of nodes, and space is O(h) for recursion stack. Mention iterative DFS with explicit stack to avoid stack overflow.

Key Points to Mention

Recursive post-order traversal to accumulate counts from children.
Base case: file node returns (1,0) if encrypted, else (0,1).
Directory node aggregates counts from all children.
Time complexity O(n), space complexity O(h) for recursion stack.
Edge cases: empty directory, null children, permission errors.
Iterative alternative using explicit stack for deep trees.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.