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IXL Learning·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Apr 2026

Summary

Got a tree problem at IXL Learning that sounds straightforward until you actually try to formalize it. The question had a nice twist that made me rethink my first instinct about how to track node values across levels.

Questions Asked (1)

Q1

Given a binary tree where each node holds a value (like a character), find the value that appears on the most distinct depth levels. For example, if 'A' shows up at level 1 and level 2, it scores 2. Describe the algorithm, argue why it's correct, and give the time and space complexity.

Algorithms & Data Structures
Author's notes

My first instinct was to just count total occurrences, which is wrong.

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AI HintsAI Generated

Suggested Approach

Clarify the problem definition, then propose a BFS traversal that tracks the set of distinct values at each depth. For each value, maintain a set of depths it appears at, and finally return the value with the largest set size. Discuss correctness, complexity, and edge cases.

Pro tip: Mention that using a set per value avoids double-counting the same value at the same depth, which is crucial for the 'distinct depth levels' requirement. Also, note that if multiple values tie, you can return any or all depending on the interviewer's preference.

1. Clarify the problem

Confirm that 'depth' starts at 0 or 1, and that we need the value appearing on the most distinct depth levels (not the most total occurrences). Ask if ties should return all values or any one.

2. Choose traversal and data structures

Use BFS (level-order traversal) to process nodes level by level. Maintain a hash map from value to a set of depths where it appears.

3. Traverse and record depths

For each node at depth d, add d to the set for its value. Use a queue to process nodes level by level, incrementing depth after each level.

4. Find the value with maximum distinct depths

After traversal, iterate through the map and track the value with the largest set size. Handle ties as clarified.

5. Analyze correctness and complexity

Argue that BFS visits each node once, and the set ensures distinct depths are counted. Time complexity is O(N) where N is number of nodes; space is O(N) for the map and queue.

Key Points to Mention

  • BFS is ideal for level-order processing and tracking depths.
  • Using a set per value prevents counting the same depth multiple times.
  • Time complexity: O(N) because each node is visited once and set operations are O(1) average.
  • Space complexity: O(N) for the queue and the map of value to depth sets.
  • Edge cases: empty tree, single node, all nodes same value, multiple values tied for max distinct depths.
  • Alternative: DFS with depth parameter also works, but BFS is more intuitive for level tracking.

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