Amazon·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Amazon SWE interview with a tree problem that looked clean on the surface but had enough edge cases to trip you up if you weren't careful. One question, but it had legs.

Questions Asked (1)

Given the root of a binary search tree and two keys p and q, find their lowest common ancestor. The LCA is the deepest node whose value falls between p and q inclusive. Implement both an iterative and a recursive solution, and handle cases where one or both keys might not exist in the tree.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The base traversal logic clicked pretty fast since BST structure makes LCA navigation straightforward.

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AI HintsAI Generated

Suggested Approach

Start by explaining the BST property that allows efficient LCA finding: traverse from the root, moving left if both keys are smaller, right if both are larger, and returning the current node when the keys split or one matches. Then implement both iterative and recursive solutions, and discuss how to handle missing keys by first verifying their existence or by modifying the traversal to track found nodes.

Pro tip: At Amazon, interviewers value candidates who proactively discuss edge cases and trade-offs. Mention that the iterative solution is more space-efficient (O(1) space) while the recursive solution is simpler but uses O(h) stack space, and always clarify assumptions about key existence before coding.

1. Clarify requirements and edge cases

Ask whether the tree can be empty, whether p and q are guaranteed to exist, and whether p and q can be equal. Discuss how to handle cases where one or both keys are missing.

2. Explain the BST LCA algorithm

Describe the standard approach: traverse from the root, moving left if both keys are less than the current node, right if both are greater, and returning the current node otherwise. This works because the LCA is the first node where the paths to p and q diverge.

3. Implement iterative solution

Write a while loop that traverses the tree, updating the current node based on comparisons with p and q. Return the current node when the split condition is met. This uses O(1) space.

4. Implement recursive solution

Write a recursive function that follows the same logic: if both keys are less, recurse left; if both are greater, recurse right; otherwise return the current node. This uses O(h) stack space.

5. Handle missing keys

Modify the algorithm to first check if both keys exist in the tree (e.g., via a search function). Alternatively, during traversal, track whether each key is found, and only return the LCA if both are present; otherwise return null or an appropriate error.

Key Points to Mention

BST property: left subtree values < node < right subtree values, enabling O(h) time complexity.
Time complexity: O(h) for both iterative and recursive solutions, where h is the height of the tree.
Space complexity: iterative uses O(1) extra space, recursive uses O(h) due to call stack.
Handling missing keys: either pre-check existence or augment traversal to track found nodes.
Edge cases: empty tree, p or q not in tree, p == q, p or q is the root.
Trade-offs: iterative is more space-efficient but slightly more code; recursive is cleaner but may cause stack overflow for skewed trees.

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