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Snowflake·Software Engineer·Technical Phone Screen·Senior

SeniorPrefer not to say
May 2026Remote

Summary

Snowflake SWE interview that went pretty deep on trees. One main coding problem but they kept pulling the thread on it, so by the end it felt like four questions in a trench coat.

Questions Asked (4)

Q1

Given a binary tree (not necessarily a BST) and two node values u and v, return the list of node values representing the shortest path from u to v. If either node is missing, return an empty list. Your solution should run in O(n) time and O(h) extra space using LCA.

Algorithms & Data Structures
Author's notes

I got the LCA idea pretty fast but fumbled on actually reconstructing the path.

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AI HintsAI Generated

Suggested Approach

Use a recursive DFS to find the lowest common ancestor (LCA) of u and v, while simultaneously checking for their existence. Then, collect the path from u to LCA and from LCA to v, and concatenate them appropriately.

Pro tip: Emphasize that the O(h) space comes from the recursion stack, and that the algorithm handles the case where one node is an ancestor of the other naturally by returning the path from the ancestor to the descendant.

1. Clarify and Plan

Confirm that the tree is not a BST, so no ordering assumptions can be made. Plan to use a single DFS traversal to find LCA and check existence.

2. Find LCA and Check Existence

Implement a recursive function that returns the LCA if both nodes are found in the subtree, or one of the nodes if only one is found, or null if neither. Use a flag or return value to indicate missing nodes.

3. Collect Paths

From the LCA, perform two separate DFS traversals to find the paths to u and v. Store the paths in lists, or use a single traversal that records the path from root to each node and then extract the segments.

4. Construct Shortest Path

Reverse the path from u to LCA (excluding LCA) and append the path from LCA to v (including LCA). This gives the shortest path from u to v.

5. Handle Edge Cases

If either u or v is missing, return an empty list. Also handle the case where u equals v, returning a list with that single value.

Key Points to Mention

  • LCA concept and its role in finding the shortest path in a tree.
  • Time complexity O(n) because each node is visited at most twice (once for LCA, once for path collection).
  • Space complexity O(h) due to recursion stack, where h is the height of the tree.
  • Handling of missing nodes: if either u or v is not present, return empty list.
  • Edge case where u and v are the same node.
  • Avoiding unnecessary traversals by combining LCA search with existence check.

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

Q2

Provide both a recursive and an iterative implementation of the path-finding solution, and walk through the time and space complexity of each.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

The recursive version came naturally.

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AI HintsAI Generated

Suggested Approach

Clarify the problem (e.g., graph traversal, grid pathfinding) and choose a concrete algorithm like DFS for recursion and BFS/DFS with an explicit stack/queue for iteration. Implement both versions cleanly, then analyze time and space complexity for each, highlighting trade-offs such as recursion depth limits and iterative overhead.

Pro tip: Mention that recursion can cause stack overflow for deep graphs, so iterative solutions are often preferred in production systems; also note that BFS guarantees shortest path in unweighted graphs while DFS does not.

1. Clarify the problem and assumptions

Ask if the graph is directed/undirected, weighted/unweighted, and whether we need any path or shortest path. Confirm the representation (adjacency list, matrix, grid).

2. Choose algorithms for each approach

For recursion, use DFS with backtracking; for iteration, use BFS with a queue or DFS with an explicit stack. Explain why you chose each.

3. Implement recursive solution

Write clean recursive code, handling base cases (found target, visited, out of bounds) and marking visited nodes to avoid cycles.

4. Implement iterative solution

Write iterative code using a stack (DFS) or queue (BFS), managing visited set and parent pointers if path reconstruction is needed.

5. Analyze time and space complexity

For both, state O(V+E) time for graph traversal. Space: recursion uses O(V) call stack; iterative uses O(V) for visited and queue/stack. Discuss worst-case and average-case.

Key Points to Mention

  • Time complexity O(V+E) for both approaches on graphs, or O(rows*cols) for grids.
  • Space complexity: recursion uses call stack O(V) in worst case; iterative uses explicit data structure O(V).
  • Recursion depth limits and stack overflow risks in production.
  • Iterative BFS guarantees shortest path in unweighted graphs; DFS does not.
  • Visited set to avoid infinite loops in cyclic graphs.
  • Trade-offs: readability vs. control, memory overhead, and tail-call optimization (if applicable).

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

Q3

If you need to answer many path queries on the same tree, what preprocessing strategies would you use? Discuss parent pointer storage, Euler tour with RMQ for LCA, and binary lifting, including the trade-offs between preprocessing cost and per-query cost.

Algorithms & Data StructuresTechnical Trade-offsSystem Design
Author's notes

This is where I started sweating.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the types of path queries (e.g., LCA, distance, k-th ancestor) and the expected query volume. Then present each preprocessing strategy (parent pointers, Euler tour + RMQ, binary lifting) with its preprocessing time, per-query time, and memory usage, and compare trade-offs. Finally, recommend a strategy based on the query mix and constraints.

Pro tip: Mention that binary lifting and Euler tour + RMQ can be combined: use binary lifting for k-th ancestor queries and Euler tour + RMQ for LCA, achieving O(1) LCA and O(log n) k-th ancestor with O(n log n) preprocessing. This shows you understand hybrid approaches.

1. Clarify query types and constraints

Ask what kinds of path queries are needed (LCA, distance, k-th ancestor, path sum, etc.) and the expected number of queries and tree size. This determines which preprocessing is optimal.

2. Explain parent pointer storage

Describe storing parent pointers for each node, enabling O(depth) traversal for LCA or ancestor queries. Preprocessing is O(n), but per-query can be O(n) in worst case (skewed tree).

3. Describe Euler tour with RMQ for LCA

Explain that an Euler tour of the tree (recording nodes on entry/exit) combined with a RMQ data structure (e.g., sparse table) allows O(1) LCA queries after O(n log n) preprocessing. Memory is O(n log n).

4. Explain binary lifting

Describe precomputing up[k][v] = 2^k-th ancestor for each node, enabling O(log n) LCA and k-th ancestor queries. Preprocessing is O(n log n) time and memory.

5. Compare trade-offs and recommend

Contrast preprocessing time/memory vs. query time: parent pointers (O(n) prep, O(n) query), Euler+RMQ (O(n log n) prep, O(1) query), binary lifting (O(n log n) prep, O(log n) query). Recommend based on query volume and type.

Key Points to Mention

  • Parent pointers: simple O(n) preprocessing, but O(n) per query in worst case; useful for small trees or few queries.
  • Euler tour + RMQ: O(n log n) preprocessing, O(1) LCA queries; requires careful implementation of Euler tour and sparse table.
  • Binary lifting: O(n log n) preprocessing, O(log n) LCA and k-th ancestor queries; memory O(n log n).
  • Trade-off: Euler+RMQ gives fastest queries but higher memory; binary lifting is more flexible for ancestor queries.
  • Hybrid approach: combine binary lifting for k-th ancestor and Euler+RMQ for LCA to get best of both.
  • Consider offline algorithms (e.g., Tarjan's offline LCA) if all queries are known in advance, achieving near-linear total time.

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

Q4

How does your solution handle edge cases: a node that doesn't exist in the tree, u equal to v, one node being an ancestor of the other, duplicate values, and very deep skewed trees where recursion might overflow the stack?

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

Duplicate values caught me off guard more than the others.

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AI HintsAI Generated

Suggested Approach

Systematically address each edge case by explaining how your solution detects and handles it, emphasizing correctness and robustness. Then discuss trade-offs, particularly for deep skewed trees, and propose iterative alternatives to recursion to avoid stack overflow. Conclude by highlighting testing strategies to ensure all edge cases are covered.

Pro tip: Mention that you would explicitly test these edge cases with unit tests and consider iterative solutions for production systems to avoid stack overflow, showing you think beyond just passing the interview.

1. Clarify the problem and assumptions

Restate the problem and confirm assumptions about the tree structure (e.g., binary tree, BST) and the definition of 'solution' (e.g., finding LCA, path sum). This ensures you address the correct edge cases.

2. Address each edge case individually

For each edge case (node not exist, u==v, ancestor relationship, duplicates, deep skewed trees), explain how your algorithm handles it, including any necessary checks or modifications.

3. Discuss trade-offs and alternatives

For deep skewed trees, compare recursive vs iterative approaches, mentioning stack overflow risks and how to mitigate them (e.g., using explicit stack, Morris traversal).

4. Highlight testing and validation

Describe how you would test these edge cases, such as writing unit tests with specific inputs and using property-based testing for duplicates.

Key Points to Mention

  • Handling non-existent nodes: return null/error or use a sentinel value, and ensure parent pointers or search logic account for absence.
  • u == v case: return u immediately or handle as a trivial case to avoid unnecessary traversal.
  • Ancestor relationship: check if one node is ancestor of the other by comparing paths or using depth information; adjust algorithm to return the ancestor.
  • Duplicate values: clarify if duplicates are allowed; if so, use node references instead of values, or define tie-breaking rules.
  • Deep skewed trees: recursion depth may cause stack overflow; propose iterative solution using explicit stack or Morris traversal to achieve O(1) space.
  • Testing: include unit tests for each edge case, and consider fuzz testing with random trees to uncover unexpected issues.

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