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Snowflake·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
May 2026

Summary

Snowflake software engineer interview with a tree manipulation problem that looked clean on paper but had a few moving parts worth thinking through carefully. The coding round felt like a mid-level systems-adjacent question dressed up as pure algorithms.

Questions Asked (2)

Q1

You have two structurally identical complete binary trees T1 and T2. Modify T2 so that each node in T2 holds the sum of all values in the corresponding subtree of T1. Implement this in O(n) time and explain your space complexity and traversal strategy.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

My first instinct was to do a preorder traversal but then I realized I needed the subtree sums, which means you actually want to compute bottom-up, so postorder made more sense.

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AI HintsAI Generated

Suggested Approach

Use a post-order traversal of T1 to compute subtree sums, storing each sum in the corresponding node of T2. Since the trees are structurally identical, you can traverse both simultaneously, achieving O(n) time and O(h) space (where h is tree height) with recursion.

Pro tip: Mention that you can avoid extra space by reusing T2's nodes to store the sums, and that an iterative post-order traversal can achieve O(1) extra space if you use parent pointers or Morris traversal.

1. Clarify the problem

Confirm that T1 and T2 are complete binary trees with identical structure, and that you need to modify T2 in-place so each node's value becomes the sum of the corresponding subtree in T1.

2. Choose traversal strategy

Select a post-order traversal (left, right, root) because subtree sums require children's sums before the parent. Traverse T1 and T2 simultaneously to map nodes.

3. Implement recursive solution

Write a recursive function that takes corresponding nodes from T1 and T2. Recursively compute left and right subtree sums, then set T2 node's value to T1 node's value plus left and right sums.

4. Analyze complexity

State that time complexity is O(n) since each node is visited once. Space complexity is O(h) for recursion stack, where h is tree height; for a complete binary tree, h = O(log n).

5. Discuss trade-offs and optimizations

Mention that an iterative post-order traversal can reduce space to O(1) if parent pointers exist, or O(h) otherwise. Also note that modifying T2 in-place avoids extra space for the result.

Key Points to Mention

  • Post-order traversal is essential because subtree sums depend on children's sums.
  • Structural identity allows simultaneous traversal of T1 and T2 without extra mapping.
  • Time complexity is O(n) as each node is processed once.
  • Space complexity is O(h) for recursion stack; for complete binary tree, h = O(log n).
  • In-place modification of T2 avoids additional space for output.
  • Iterative approaches (e.g., using explicit stack or Morris traversal) can reduce space complexity.

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

Q2

Implement a buildCompleteTree function that takes a level-order integer array and constructs a complete binary tree from it.

Algorithms & Data Structures
Author's notes

Pretty standard BFS construction, use a queue and assign left/right children as you pop nodes.

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AI HintsAI Generated

Suggested Approach

Clarify the problem constraints and edge cases, then explain that a complete binary tree can be built by mapping array indices to tree nodes using the relationship: for index i, left child is at 2i+1 and right child at 2i+2. Implement the construction iteratively using a queue or recursively, ensuring all nodes are linked correctly.

Pro tip: Mention that this index mapping works because a complete binary tree fills levels left-to-right, so the array representation is compact with no gaps. Also, discuss handling of null or empty input and potential integer overflow for large indices.

1. Understand the problem

Restate the problem: given a level-order array, construct a complete binary tree. Clarify that the array represents a complete tree, so no null markers are needed except possibly for empty input.

2. Choose an approach

Decide between iterative (using a queue) or recursive (using index mapping) construction. Explain the trade-offs: iterative is intuitive, recursive is concise but may risk stack overflow for deep trees.

3. Implement the construction

For iterative: create root from first element, use a queue to track parent nodes, and assign left and right children from subsequent array elements. For recursive: define a helper that takes an index, creates a node if index is within bounds, and recursively builds left and right subtrees.

4. Test with examples

Walk through a small example (e.g., [1,2,3,4,5]) to verify the tree structure. Check edge cases: empty array, single element, and array with only left children.

5. Analyze complexity

State that both approaches run in O(n) time and O(n) space (for the tree and queue/recursion stack). Mention that the recursive approach uses O(h) stack space, where h is the height.

Key Points to Mention

  • Definition of a complete binary tree: all levels filled except possibly the last, which is filled left to right.
  • Index mapping: for node at index i, left child at 2i+1, right child at 2i+2.
  • Iterative BFS approach using a queue to assign children level by level.
  • Recursive approach using index bounds to avoid null checks.
  • Handling edge cases: empty array, single node, and arrays with missing children.
  • Time and space complexity: O(n) time, O(n) space for the tree, and O(n) or O(h) auxiliary space depending on approach.

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