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Databricks·Software Engineer·Technical Phone Screen·Senior

Senior
May 2026

Summary

Databricks SWE interview with a pretty gnarly tree problem that you absolutely cannot brute force your way through. The recursive structure is the whole point and if you miss that you're cooked.

Questions Asked (1)

Q1

A Fibonacci tree T(k) is defined recursively where T(1) and T(2) are single nodes, and for k >= 3 the root has T(k-1) as its left subtree and T(k-2) as its right. Nodes are labeled 1-based in preorder. Given k up to 60 and two node labels a and b, return the sequence of node labels along the path from a to b.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

k up to 60 was the thing that should've told me immediately: you can't build this tree.

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AI HintsAI Generated

Suggested Approach

First, explain how to navigate the Fibonacci tree without building it, using the recursive structure and subtree sizes to map a label to its path from the root. Then, find the lowest common ancestor (LCA) of the two nodes by comparing their root-to-node paths, and finally concatenate the path from a up to the LCA and from the LCA down to b.

Pro tip: Mention that the tree has exponential size (Fibonacci number of nodes), so any solution must work in O(k) time and space by exploiting the recursive definition and not constructing the tree. Also, note that preorder labeling allows determining which subtree a node belongs to by comparing its label with the size of the left subtree.

1. Understand the tree structure and labeling

Explain that T(k) has size F(k) (Fibonacci number), with left subtree T(k-1) and right subtree T(k-2). In preorder, the root is label 1, the left subtree contains labels 2 to F(k-1)+1, and the right subtree contains the rest.

2. Compute subtree sizes efficiently

Precompute Fibonacci numbers up to k=60 (using 64-bit integers) to quickly get the size of any subtree T(i). This avoids exponential traversal.

3. Find path from root to a given node

Starting from the root of T(k), recursively determine whether the target label lies in the left or right subtree by comparing with the left subtree size. Record the labels along the path.

4. Find the lowest common ancestor (LCA)

Compute the root-to-node paths for both a and b, then find the last common node in these paths. This is the LCA.

5. Construct the final path

The path from a to b is the reverse of the path from a to LCA (excluding LCA) followed by the path from LCA to b (including LCA). Return this sequence.

Key Points to Mention

  • The tree size grows exponentially (Fibonacci), so we cannot build it; we must use the recursive structure.
  • Preorder labeling: root is 1, left subtree labels are 2 to size(left)+1, right subtree labels are size(left)+2 to size(T(k)).
  • Use Fibonacci numbers to compute subtree sizes in O(1) after O(k) precomputation.
  • Finding the path from root to a node takes O(k) time by recursively descending.
  • The LCA can be found by comparing the two root-to-node paths; the last common node is the LCA.
  • The final path is constructed by concatenating the upward path from a to LCA (excluding LCA) and the downward path from LCA to b.

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