Meta·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jun 2026Remote

Summary

Meta SWE interview with a tree problem that looks manageable until you realize the edge cases are where they actually want to see you think. Single coding round, no behavioral from what I can tell.

Questions Asked (1)

Given an N-ary tree and two node IDs, find and return their Lowest Common Ancestor. If either node doesn't exist in the tree, return null.

Algorithms & Data Structures

Author's notes

My first instinct was to just do two root-to-node path traversals and find where they diverge, which works fine.

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AI HintsAI Generated

Suggested Approach

Use a post-order DFS that returns the node if it matches either target, otherwise recursively searches children. At each node, if two or more children return non-null results (or one child plus the node itself matches), the current node is the LCA. If only one child returns a result, propagate it upward; if none, return null.

Pro tip: Clarify edge cases upfront: whether the tree is static, if node IDs are unique, and if a node can be an ancestor of itself. This shows thoroughness and avoids incorrect assumptions.

1. Clarify requirements and edge cases

Ask about tree size, node ID uniqueness, and whether a node can be its own ancestor. Confirm return type (node object vs. ID) and handling of missing nodes.

2. Choose traversal strategy

Select a recursive post-order DFS because it naturally processes children before the parent, allowing LCA detection when results from multiple subtrees combine.

3. Define recursive function

The function returns the LCA if found, or the target node if only one is found in the subtree, or null otherwise. At each node, check if it matches a target and recursively process all children.

4. Combine results and detect LCA

Count how many children return non-null. If the current node matches a target, increment count. If count >= 2, return current node as LCA; if count == 1, return the non-null result; else return null.

5. Handle missing nodes and verify

After traversal, if the result is not the LCA (e.g., only one target found), return null. Walk through examples and discuss time/space complexity.

Key Points to Mention

Time complexity O(N) where N is number of nodes, as each node is visited once.
Space complexity O(H) for recursion stack, where H is tree height; worst case O(N) for skewed tree.
Post-order traversal ensures children are processed before parent, enabling LCA detection.
Handling of edge cases: one or both nodes missing, root as LCA, node being its own ancestor.
Use of a helper function that returns a node (LCA or target) and null otherwise.
Potential optimization: early termination if both nodes found, but not necessary for O(N).

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