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Databricks·Software Engineer·Onsite - Coding / Algorithms·Senior

Senior
May 2026

Summary

Databricks SWE interview threw a pretty gnarly tree problem at me. The whole thing revolves around Fibonacci trees and navigating them implicitly since the tree can have trillions of nodes. Solid algorithmic challenge, probably onsite-level difficulty.

Questions Asked (1)

Q1

Given a Fibonacci tree of order k (defined recursively where T(k) has T(k-1) as left subtree and T(k-2) as right subtree), nodes are numbered in preorder from 1 to size(k). For two given node indices a and b, return the sequence of preorder indices along the simple path from a to b. The tree can be astronomically large so you cannot build it explicitly.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

This one took me a while to even parse.

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AI HintsAI Generated

Suggested Approach

Recognize that the preorder numbering of a Fibonacci tree follows a recursive pattern based on subtree sizes. Use the recurrence size(k) = size(k-1) + size(k-2) + 1 to determine which subtree a node belongs to and its relative index, then compute the path by finding the lowest common ancestor (LCA) and concatenating the upward and downward paths. Since the tree is huge, avoid explicit construction and rely on arithmetic and recursion.

Pro tip: Emphasize that the problem reduces to finding the LCA in a recursively defined tree, and that the preorder indices can be computed on the fly using the Fibonacci-like size recurrence. Mention that this approach runs in O(k) time, which is logarithmic in the tree size, making it efficient for astronomical trees.

1. Understand the tree structure and preorder numbering

Explain that T(k) has left subtree T(k-1) and right subtree T(k-2), and that preorder numbering assigns the root index 1, then numbers the left subtree, then the right subtree. Derive the size recurrence: size(0)=0, size(1)=1, size(k)=size(k-1)+size(k-2)+1.

2. Determine subtree membership for a given index

Given a node index x in T(k), if x=1 it's the root; if 2 ≤ x ≤ size(k-1)+1, it's in the left subtree with relative index x-1; otherwise it's in the right subtree with relative index x - size(k-1) - 1. Use this to recursively locate the node.

3. Compute the path from root to a node

Recursively build the path from the root of T(k) to the target node by recording the current root index and descending into the appropriate subtree, adjusting the index accordingly. This yields a sequence of preorder indices from root to node.

4. Find the lowest common ancestor (LCA) of a and b

Use the root-to-node paths to find the last common index, which is the LCA. Alternatively, recursively determine the LCA by comparing which subtrees contain a and b.

5. Construct the full path from a to b

Take the path from a up to the LCA (excluding LCA) and concatenate with the reverse of the path from LCA down to b (including LCA). Ensure the order is correct: from a upwards to LCA, then downwards to b.

Key Points to Mention

  • The size recurrence size(k) = size(k-1) + size(k-2) + 1 and its closed form related to Fibonacci numbers.
  • How to map a global preorder index to a local index within a subtree using size(k-1) and size(k-2).
  • The concept of lowest common ancestor (LCA) and its role in path finding.
  • Time complexity: O(k) for path computation, which is O(log N) where N is the number of nodes.
  • Space complexity: O(k) for storing the paths, which is efficient.
  • Handling edge cases: a = b, a or b is the root, and very large k (up to 10^9) requiring iterative or memoized recursion.

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