Graph Valid Tree
Given $n$ nodes labeled from 0 to $n-1$ and a list of undirected edges, write a function to check whether these edges form a valid tree. Use Union-Find or BFS/DFS.
Why Interviewers Ask This
Interviewers at LinkedIn ask this to assess your ability to model real-world connectivity problems, such as social network dependencies. They evaluate whether you can distinguish between a tree and a general graph by identifying cycles or disconnected components. This tests your grasp of fundamental data structures like Union-Find or traversal algorithms under constraints.
How to Answer This Question
1. Clarify the definition: A valid tree must have exactly n-1 edges, be fully connected (one component), and contain no cycles. 2. Choose your strategy: Opt for Union-Find for O(E alpha(N)) efficiency or DFS/BFS for O(V+E) complexity. 3. Validate edge count first: If edges != n-1, immediately return false to save computation time. 4. Implement cycle detection: During traversal or union operations, check if two nodes are already in the same set or visited. 5. Verify connectivity: Ensure all nodes from 0 to n-1 are reachable from a single root node after processing edges. 6. Handle edge cases: Explicitly address scenarios where n=0 or n=1, and ensure the input list is treated as undirected.
Key Points to Cover
- Explicitly stating the necessary condition that edges must equal n-1
- Choosing Union-Find for its dual capability in cycle and connectivity detection
- Demonstrating awareness of time complexity differences between approaches
- Handling edge cases like n=0 or n=1 gracefully
- Explaining the logic clearly rather than just writing code
Sample Answer
To solve this efficiently, I would first validate the structural prerequisites. A valid tree with n nodes must have exactly n-1 edges. If the input list doesn't match this count, it's impossible to form a tree without either a cycle or a disconnected component, so we can return false immediately. Next, I'd use the Union-Find algorithm because it elegantly handles both cycle detection and connectivity checks in near-linear time. I'll initialize each node as its own parent. As I iterate through the edges, I'll find the root of both nodes involved. If they share the same root, adding this edge creates a cycle, which invalidates the tree structure. If their roots differ, I union them. After processing all edges, I must also ensure that there is only one connected component. With Union-Find, if we successfully processed n-1 edges without finding a cycle, connectivity is implicitly guaranteed. However, if the edge count wasn't pre-checked, I would verify that all nodes eventually point to the same root. For example, with 4 nodes and edges [[0,1], [1,2], [2,3]], the algorithm unions them sequentially. Since no cycles occur and we end with one component, it returns true. Conversely, adding an edge [0,3] would detect a cycle, returning false. This approach mirrors how LinkedIn might optimize dependency resolution in large-scale systems.
Common Mistakes to Avoid
- Failing to check the edge count before running complex algorithms, wasting time on obvious failures
- Ignoring the undirected nature of edges, leading to incorrect cycle detection logic
- Missing the requirement that the graph must be fully connected, not just acyclic
- Not handling the base case where n equals 0 or 1 correctly
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