Meta·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Apr 2026

Summary

Meta SWE interview with a tree traversal problem that had a twist baked in. Pretty focused session, one meaty coding question with follow-ups on complexity and design choices.

Questions Asked (1)

Given a binary tree and a predicate function P that evaluates node values, return a list of lists where each inner list holds the values at a given depth that satisfy P. You must implement this iteratively using a stack, no recursion allowed. Explain how you track level boundaries, manage per-level results, and give a complexity analysis.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The predicate filter part was fine, the annoying bit was detecting level boundaries without recursion.

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AI HintsAI Generated

Suggested Approach

Use an iterative depth-first search with a stack that stores nodes along with their depth. Track level boundaries by comparing the depth of the current node with the previous node's depth, starting a new inner list when the depth increases. Apply the predicate P to each node's value and append to the current level's list if it satisfies P.

Pro tip: Clarify whether the output should include empty lists for levels with no matching values; handling this edge case shows attention to detail. Also, mention that using a stack (LIFO) means you'll traverse right-to-left if you push left then right, which is fine for this problem but worth noting.

1. Clarify requirements and edge cases

Confirm the output format (list of lists, one per depth, possibly empty), and discuss edge cases like empty tree, predicate always false, or skewed tree.

2. Design stack-based traversal

Use a stack of (node, depth) pairs. Initialize with (root, 0). While stack is not empty, pop a node, and process it based on its depth.

3. Track level boundaries

Maintain a variable for the current depth and a list for the current level. When the popped node's depth is greater than the current depth, start a new level list and update current depth.

4. Apply predicate and collect results

If the node's value satisfies P, append it to the current level's list. Push the node's children with depth+1 onto the stack (order doesn't matter for correctness, but note traversal order).

5. Analyze complexity and discuss trade-offs

State time complexity O(n) and space complexity O(n) in worst case. Mention that stack-based DFS uses O(h) space for balanced trees but O(n) for skewed trees, and compare with BFS queue approach.

Key Points to Mention

Stack stores both node and depth to track level boundaries without recursion.
Level boundaries are detected when the depth of the current node exceeds the last recorded depth.
Predicate P is applied to each node's value; only matching values are added to the level list.
Time complexity is O(n) since each node is visited once; space complexity is O(n) worst-case due to stack size.
Edge cases: empty tree returns empty list; levels with no matches may be empty lists or omitted (clarify).
Iterative DFS with stack vs BFS with queue: both work, but stack may reverse order of levels if not careful.

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