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Snowflake·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
May 2026Remote

Summary

Snowflake SWE coding round, basically one meaty tree problem that took up the whole session. The core question is close enough to LC 545 that if you've seen it you'll recognize it, but the edge cases are where things get tricky and they pushed hard on follow-ups.

Questions Asked (2)

Q1

Given the root of a binary tree, compute the sum of the left boundary nodes, the right boundary nodes, and all leaf nodes, without double-counting any node that appears in multiple categories.

Algorithms & Data Structures
Author's notes

I recognized the structure pretty fast since I'd done the boundary traversal problem before, but I still fumbled the edge cases.

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AI HintsAI Generated

Suggested Approach

Break the problem into three separate traversals: one for the left boundary (excluding leaves), one for the right boundary (excluding leaves), and one for all leaves. Use a set to track visited nodes to avoid double-counting, or carefully define traversal rules to naturally exclude overlaps. Then sum the values from all three collections.

Pro tip: Clarify the definition of 'boundary' upfront—especially whether the root counts as both left and right boundary when the tree has only one child—and mention that you'll handle edge cases like a single-node tree or skewed trees explicitly.

1. Clarify definitions and edge cases

Confirm what constitutes left boundary, right boundary, and leaves, and how to handle overlaps (e.g., root, single-child nodes). Discuss edge cases like empty tree, single node, and skewed trees.

2. Design traversal strategy

Plan three separate traversals: left boundary (top-down, excluding leaves), right boundary (bottom-up, excluding leaves), and leaf collection (any traversal). Use a set to track visited nodes if needed.

3. Implement boundary collection

Write helper functions for each traversal, ensuring leaves are excluded from boundary traversals and that the root is handled correctly (e.g., add root once).

4. Combine and sum

Merge the three collections, remove duplicates (if not already handled), and compute the sum. Return the result.

5. Analyze complexity and test

State time and space complexity (O(n) time, O(n) space for set/recursion). Walk through test cases to verify correctness.

Key Points to Mention

  • Definition of boundary: left boundary excludes leaves, right boundary excludes leaves, leaves are all leaf nodes.
  • Handling overlaps: root may be counted once; use a set or careful traversal to avoid double-counting.
  • Traversal order: left boundary top-down, right boundary bottom-up, leaves in any order.
  • Edge cases: empty tree, single node, tree with only left or right children, skewed trees.
  • Time and space complexity: O(n) time, O(n) space due to recursion stack and set.
  • Alternative approaches: iterative traversal with explicit stack, or single traversal with flags.

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

Q2

How would you extend the solution to return the actual boundary node values in order, handle very deep trees without recursion, or generalize the approach to trees where nodes can have more than two children?

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

The iterative version for deep trees is where I started to lose steam.

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AI HintsAI Generated

Suggested Approach

Treat the question as three separate extensions: (1) modify the traversal to collect boundary nodes in order, (2) replace recursion with an explicit stack or iterative traversal to handle deep trees, and (3) generalize the traversal logic to n-ary trees by iterating over all children. For each, explain the algorithmic changes, complexity implications, and trade-offs.

Pro tip: Emphasize that avoiding recursion is crucial for deep trees to prevent stack overflow, and that generalizing to n-ary trees often simplifies the code because you no longer assume left/right children. Also, mention that collecting boundary nodes in order requires careful handling of duplicates and edge cases like single-node trees.

1. Clarify the extensions

Restate the three sub-problems to ensure you understand: returning boundary values in order, handling deep trees without recursion, and supporting n-ary trees. Ask clarifying questions if needed, such as whether the boundary should include leaves and how to handle duplicates.

2. Extend to return boundary values

Describe how to modify the traversal to collect nodes: typically a combination of left boundary (excluding leaves), leaves, and right boundary (excluding leaves) in reverse order. Discuss using a list to accumulate values and ensuring correct order.

3. Convert to iterative approach

Explain how to replace recursion with an explicit stack (or queue) for each traversal phase. For deep trees, this avoids stack overflow and gives control over memory usage. Mention that iterative traversals can be more complex but are necessary for robustness.

4. Generalize to n-ary trees

Show how to adapt the algorithm to nodes with multiple children. For boundary traversal, define left boundary as the first child at each level, right boundary as the last child, and leaves as nodes with no children. Iterate over children lists instead of left/right pointers.

5. Analyze trade-offs and complexity

Discuss time and space complexity for each extension. Iterative approaches may use O(h) space for the stack, where h is height, but avoid recursion limits. N-ary generalization may increase branching factor but simplifies logic. Mention potential edge cases and testing strategies.

Key Points to Mention

  • Boundary traversal definition: left boundary (top-down, excluding leaves), leaves (left-to-right), right boundary (bottom-up, excluding leaves).
  • Iterative traversal using explicit stack to avoid recursion depth limits, especially for skewed trees.
  • Handling n-ary trees: iterate over children list, define leftmost and rightmost paths, and collect leaves.
  • Time complexity: O(N) for all nodes; space complexity: O(H) for iterative stack, where H is tree height.
  • Edge cases: single node, only left/right children, duplicate values, and deep trees causing stack overflow.
  • Trade-offs: iterative code is more verbose but safer; recursive is simpler but limited by stack depth.

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