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Amazon·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jun 2026

Summary

Amazon SWE interview focused on a classic tree problem but pushed well beyond the surface, asking about stack overflow handling and complexity tradeoffs. The depth of follow-up questions made it feel more like a system design conversation than a pure coding round.

Questions Asked (3)

Q1

Given a potentially very large binary tree, compute its diameter (the number of edges on the longest path between any two nodes) in O(n) time. Walk through your recursive or iterative approach and justify why it's correct.

Algorithms & Data Structures
Author's notes

My first instinct was to write the naive two-pass solution where you compute height separately for each node, which is obviously O(n^2).

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AI HintsAI Generated

Suggested Approach

Use a post-order traversal to compute the height of each subtree while simultaneously tracking the maximum diameter seen so far. At each node, the diameter through that node is the sum of the heights of its left and right subtrees; update the global maximum accordingly. Return the height to the parent to enable O(n) time.

Pro tip: Emphasize that the diameter may not pass through the root, so you must consider every node as a potential 'highest' point. Also, mention that the algorithm runs in O(n) time and O(h) space due to recursion, and discuss iterative approaches if recursion depth is a concern.

1. Clarify the problem and constraints

Confirm that diameter is measured in edges, not nodes, and that the tree can be very large, so an O(n) solution is required. Discuss potential recursion depth issues and whether an iterative solution is preferred.

2. Define the recursive function

Define a function that returns the height of a subtree (max edges from root to leaf) and updates a global variable tracking the maximum diameter. At each node, compute left and right heights recursively.

3. Compute diameter at each node

The diameter passing through the current node is left_height + right_height. Update the global maximum if this sum is larger. Then return 1 + max(left_height, right_height) as the height to the parent.

4. Handle base cases and edge cases

For a null node, return height -1 (if counting edges) or 0 (if counting nodes) and ensure the global maximum is initialized appropriately. Consider a single-node tree (diameter 0) and skewed trees.

5. Analyze complexity and correctness

Explain that each node is visited once, so time is O(n). Space is O(h) for recursion stack, where h is tree height. Justify correctness by noting that the longest path must have a highest node, and the algorithm considers all such nodes.

Key Points to Mention

  • Post-order traversal to compute heights bottom-up
  • Global variable to track maximum diameter
  • Diameter through a node = left height + right height
  • Height of a node = 1 + max(left height, right height)
  • Time complexity O(n), space complexity O(h) due to recursion
  • Iterative approach using stack for post-order traversal to avoid recursion depth issues

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

Q2

How would you handle very deep trees where recursion could cause a stack overflow? What's your iterative alternative?

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

This is where I got a little shaky.

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AI HintsAI Generated

Suggested Approach

Start by acknowledging the problem: recursion depth is limited by the call stack, so deep trees can cause stack overflow. Then explain that an iterative solution using an explicit stack (or queue for BFS) avoids this by moving the state to the heap, which is much larger. Finally, discuss trade-offs like increased code complexity and memory usage, and mention when recursion might still be acceptable (e.g., balanced trees).

Pro tip: Mention that you can also increase the stack size or use tail recursion in languages that optimize it, but emphasize that iterative is generally more robust and portable. Also, relate this to Amazon's leadership principles by highlighting customer obsession (reliability) and dive deep (understanding memory models).

1. Identify the problem

Explain that recursion uses the call stack, which has limited size, and deep trees can cause stack overflow. Mention that this is especially critical in production systems where reliability is key.

2. Propose iterative alternative

Describe using an explicit stack (for DFS) or queue (for BFS) to simulate the recursion. Show how to push child nodes onto the stack and process them in a loop.

3. Discuss trade-offs

Compare recursion vs iteration: recursion is often simpler and more readable, but iterative avoids stack overflow and can be more memory-efficient for deep trees. Mention that iterative may require more code and careful handling of state.

4. Consider optimizations

Mention techniques like Morris traversal for O(1) space, or using a hybrid approach (e.g., recursion for shallow parts, iteration for deep parts). Also note that some languages support tail call optimization.

5. Relate to real-world scenarios

Give an example where deep trees occur (e.g., parsing deeply nested JSON, file systems) and explain how you would choose the approach based on constraints like memory, performance, and code maintainability.

Key Points to Mention

  • Stack overflow due to limited call stack size
  • Explicit stack/queue for iterative traversal
  • Heap vs stack memory allocation
  • Trade-offs: readability vs robustness, memory usage
  • Morris traversal for O(1) space
  • Tail recursion optimization (if applicable)

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

Q3

What are the time and space complexities of your solution, and how does the space complexity change between the recursive and iterative versions?

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

Time is O(n) either way, that part was easy.

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AI HintsAI Generated

Suggested Approach

Start by clearly stating the time and space complexity of your solution, then compare the recursive and iterative versions, focusing on how recursion uses the call stack while iteration uses explicit data structures. Explain the trade-offs and justify your choice based on the problem constraints and environment.

Pro tip: At Amazon, interviewers value candidates who not only state complexities but also discuss practical implications, such as stack overflow risks in recursion or memory overhead in iterative approaches, and how these affect scalability.

1. State the time complexity

Clearly specify the time complexity of your solution, e.g., O(n) or O(n log n), and briefly explain why by analyzing the number of operations relative to input size.

2. State the space complexity

Specify the space complexity, including auxiliary space, and identify the main contributors such as recursion stack, explicit data structures, or input storage.

3. Compare recursive vs. iterative space

Explain that recursion adds O(d) space for the call stack (d = depth), while iteration may use O(1) extra space if no data structure is used, or O(n) if a stack/queue is used.

4. Discuss trade-offs and optimizations

Mention trade-offs like code simplicity vs. memory usage, and potential optimizations such as tail recursion or converting to iteration to reduce space.

5. Relate to problem constraints

Tie the analysis back to the problem's constraints and the production environment, explaining why one approach might be preferred for scalability or readability.

Key Points to Mention

  • Time complexity analysis based on input size and operations
  • Space complexity includes auxiliary space and input space
  • Recursion uses call stack space proportional to depth
  • Iteration can achieve O(1) space if no extra data structures are used
  • Trade-offs between readability, performance, and memory
  • Potential stack overflow in recursion for large inputs

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