Bytedance·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jun 2026Remote

Summary

Bytedance coding round, graph problem that looks straightforward until you realize there are two separate things to check. Not a bad experience, just a lot riding on one question.

Questions Asked (1)

Given n nodes labeled 0 to n-1 and a list of undirected edges, determine whether the graph forms a valid tree.

Algorithms & Data Structures

Author's notes

I went with Union-Find because I've drilled it enough that it feels safer under pressure.

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AI HintsAI Generated

Suggested Approach

Clarify that a valid tree must be connected and acyclic, then use Union-Find to detect cycles while counting edges, or BFS/DFS to check connectivity and cycle presence. Start by stating the two necessary conditions: exactly n-1 edges and no cycles (or full connectivity).

Pro tip: Mention that if the number of edges is not exactly n-1, you can immediately return false, which is a quick optimization and shows you understand the mathematical property of trees. Also, discuss trade-offs between Union-Find and BFS/DFS in terms of time and space complexity.

1. Clarify the definition of a valid tree

Confirm that a valid tree is a connected acyclic undirected graph with exactly n-1 edges. State that if the graph has fewer or more than n-1 edges, it cannot be a tree.

2. Choose an algorithm

Select either Union-Find (Disjoint Set Union) for cycle detection or BFS/DFS for connectivity and cycle checking. Explain why the chosen method is suitable.

3. Outline the algorithm steps

For Union-Find: initialize parent array, iterate edges, union nodes, and if an edge connects nodes already in the same set, a cycle exists. For BFS/DFS: build adjacency list, traverse from node 0, check if all nodes visited and no back edges (except parent).

4. Analyze complexity and edge cases

State time and space complexity: O(n α(n)) for Union-Find, O(n + e) for BFS/DFS. Mention edge cases: n=1 (single node is a tree), disconnected graph, self-loops, and multiple edges.

5. Conclude with validation

Summarize that the graph is a valid tree if and only if it has exactly n-1 edges and is connected (or acyclic). Optionally, mention that you can return early if edge count condition fails.

Key Points to Mention

A tree must have exactly n-1 edges and be connected (or acyclic).
Union-Find can detect cycles efficiently by checking if two nodes are already in the same set before union.
BFS/DFS can verify connectivity and detect cycles by tracking visited nodes and parent pointers.
Time complexity: O(n + e) for BFS/DFS, O(n α(n)) for Union-Find, where α is the inverse Ackermann function.
Edge cases: n=1 (valid tree), disconnected components, self-loops, and multiple edges between same nodes.
Early termination: if edge count != n-1, return false immediately.

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