← Oracle Interview Insights

Oracle·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jun 2026

Summary

Oracle SWE interview with a tree problem that sounds easy until you actually have to code it up and explain the complexity properly. Pretty standard technical screen vibe.

Questions Asked (1)

Q1

Given two N-ary trees, write a function to check if they are identical. Two trees are identical if their root values match and their children lists are recursively identical in the same order. Also discuss time and space complexity.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

The recursive solution isn't hard to sketch out but I fumbled a bit explaining the complexity.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Start by clarifying the problem and edge cases, then propose a recursive solution that simultaneously traverses both trees, comparing root values and recursively checking each child pair in order. After outlining the algorithm, analyze time and space complexity, and discuss potential optimizations or trade-offs.

Pro tip: Mention that you can short-circuit on the first mismatch to save time, and that the space complexity is O(h) for recursion depth, but can be O(N) in the worst case for skewed trees. Also, note that an iterative approach using a stack can avoid recursion limits.

1. Clarify and Define

Restate the problem in your own words, confirm the definition of identical N-ary trees, and ask about constraints (e.g., tree size, recursion depth limits).

2. Outline Recursive Approach

Explain that you will recursively compare the roots and then iterate through both children lists simultaneously, ensuring same length and recursive equality.

3. Handle Edge Cases

Discuss base cases: both null, one null, different values, different number of children. Also consider empty trees and single-node trees.

4. Analyze Complexity

State that time complexity is O(min(N, M)) where N and M are the number of nodes in each tree, as you stop at first mismatch. Space complexity is O(min(H1, H2)) for recursion stack, where H is height.

5. Discuss Trade-offs and Optimizations

Mention iterative alternative using stack/queue, potential for early termination, and that the recursive solution is clean but may risk stack overflow for deep trees.

Key Points to Mention

  • Recursive comparison of root values and children lists in order.
  • Base cases: both null, one null, value mismatch, child count mismatch.
  • Time complexity: O(min(N, M)) due to early termination on mismatch.
  • Space complexity: O(min(H1, H2)) for recursion stack, worst-case O(N) for skewed trees.
  • Iterative alternative using explicit stack to avoid recursion depth issues.
  • Short-circuit evaluation to improve average-case performance.

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