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Databricks·Software Engineer·Technical Phone Screen·Senior

Senior
Jun 2026

Summary

Databricks software engineering interview with a tree path problem that looks deceptively clean on the surface but gets tricky fast once you realize you can't just build the tree explicitly.

Questions Asked (1)

Q1

You have a recursively defined binary tree where T(k) has a root, a left subtree T(k-1), and a right subtree T(k-2), with base cases T(0) and T(1) each being a single node. Nodes are labeled 0 to N-1 in preorder. Given an order, a start node, and an end node, return the unique path between them using only arithmetic, no explicit tree construction.

Algorithms & Data Structures
Author's notes

The preorder labeling part clicked pretty quickly for me, size(k) follows a Fibonacci-like recurrence so you can compute any subtree's size and figure out whether a given label lives in the left or right subtree.

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AI HintsAI Generated

Suggested Approach

First, recognize that the tree structure is determined by Fibonacci-like recursion, so node indices in preorder can be mapped to tree positions using Fibonacci numbers. Then, compute the path by finding the lowest common ancestor (LCA) using arithmetic on indices, and construct the path from start to LCA and LCA to end.

Pro tip: Precompute Fibonacci numbers up to N to quickly determine subtree sizes and navigate the tree without explicit construction. Also, handle edge cases like when start equals end or when one node is an ancestor of the other.

1. Understand the tree structure and labeling

The tree T(k) has size F(k+2)-1 where F is Fibonacci (with F(0)=0, F(1)=1). Nodes are labeled in preorder: root first, then left subtree, then right subtree.

2. Map node index to its position in the tree

Given a node index, determine its depth and the subtree it belongs to by comparing with Fibonacci numbers representing subtree sizes.

3. Find the lowest common ancestor (LCA)

Using the depth and subtree information, find the LCA of the start and end nodes by moving up the tree until both nodes are in the same subtree.

4. Construct the path

From start node, move up to LCA, then down to end node, using arithmetic to compute parent and child indices based on preorder labeling.

5. Handle edge cases and verify

Check if start equals end, if one is ancestor of the other, and ensure the path is correct by testing with small examples.

Key Points to Mention

  • Fibonacci numbers and their relation to subtree sizes
  • Preorder traversal and node labeling
  • Lowest common ancestor (LCA) concept
  • Arithmetic computation of parent and child indices
  • Time and space complexity: O(log N) time, O(1) space with precomputed Fibonacci numbers
  • Edge cases: start equals end, ancestor-descendant relationship

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