Atlassian·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jun 2026Remote

Summary

Atlassian coding round, tree problem that looked straightforward until I started second-guessing my traversal approach mid-interview. Not my cleanest performance but I got through it.

Questions Asked (1)

Given an organization modeled as a tree (each employee belongs to exactly one org, each org has at most one parent), find the lowest common ancestor of two given employees. Walk through your data structure choice, traversal strategy, and complexity.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I went with the set-of-ancestors approach first: walk up from one node collecting all ancestors, then walk up from the other and return the first match.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the tree structure (rooted, parent pointers, etc.) and then present a solution using parent pointers and a hash set to find the LCA in O(h) time and O(h) space. Alternatively, discuss binary lifting for O(log n) query time after O(n log n) preprocessing, and compare trade-offs.

Pro tip: Mention that in a real org chart, the tree might be deep and unbalanced, so binary lifting or Euler tour + RMQ can be more efficient for repeated queries. Also, consider if the employees are in different orgs (no common ancestor) and handle that edge case.

1. Clarify the problem and assumptions

Ask if the tree is rooted, if nodes have parent pointers, if the tree is static, and if multiple queries will be made. Confirm that each employee belongs to exactly one org and each org has at most one parent.

2. Choose data structure and traversal strategy

For a single query, use parent pointers and a hash set: traverse from one node to root, storing visited nodes, then traverse from the other node until a common node is found. For multiple queries, consider binary lifting or Euler tour + RMQ.

3. Walk through the algorithm

Explain step-by-step: start from employee A, go up to root, add each node to a set. Then start from employee B, go up until you find a node in the set. That node is the LCA. Handle cases where one node is ancestor of the other.

4. Analyze complexity

For the hash set approach: O(h) time and O(h) space, where h is the height of the tree. For binary lifting: O(n log n) preprocessing, O(log n) per query, O(n log n) space. Discuss trade-offs based on number of queries.

5. Discuss edge cases and optimizations

Mention edge cases: same employee, one is ancestor of the other, employees in different trees (no LCA). Optimizations: if parent pointers not available, build them via DFS; use binary lifting for many queries.

Key Points to Mention

Parent pointers allow upward traversal; if not present, build them with DFS/BFS.
Hash set approach for single query: O(h) time, O(h) space.
Binary lifting for multiple queries: O(n log n) preprocessing, O(log n) per query.
Euler tour + RMQ (sparse table) for O(1) query after O(n log n) preprocessing.
Edge cases: same node, ancestor relationship, disconnected trees.
Trade-offs: preprocessing vs query time, memory usage, tree depth.

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