Compute binary-tree diameter via return-only DFS
Company: Meta
Role: Data Scientist
Category: Coding & Algorithms
Difficulty: medium
Interview Round: Technical Screen
Given the root of a binary tree, compute its diameter defined as the number of edges on the longest path between any two nodes. Implement a DFS that returns a tuple to its caller: (height_of_subtree, best_diameter_in_subtree). Combine children’s return values at each node to update the best diameter without using any external list, array, or global variable. Requirements: O(n) time, O(h) auxiliary space where h is the tree height. Edge cases: empty tree (diameter = 0), single node (diameter = 0), completely skewed tree. Then answer: (1) Why does accumulating traversal state in a list during DFS inflate space from O(h) to O(n), and under what input shapes does it hurt most? (2) Provide an iterative postorder version using an explicit stack that still achieves O(h) auxiliary space. (3) State and justify the time and space complexities for both versions.
Quick Answer: This question evaluates algorithmic problem-solving with binary trees, focusing on tree traversal techniques, recursion versus iterative implementations, and analysis of time and space complexity including worst-case input shapes such as skewed trees.
Given level-order tree values, return the diameter in edges using DFS that returns height and best diameter.
Constraints
- Inputs are provided as Python literals matching the function signature.
- Return a deterministic exact-match result.
Examples
Input: ([1,2,3,4,5],)
Expected Output: 3
Explanation: Diameter through root.
Input: ([],)
Expected Output: 0
Explanation: Empty tree.
Hints
- Choose a representation that makes the core operation simple.
- Handle empty and boundary inputs before the main algorithm.