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🌳 Graphs: Graph Valid Tree

📝 Problem Description

LeetCode 261: Graph Valid Tree

Given n nodes labeled from 0 to n - 1 and a list of edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.

Real-World Application

Network topology management, directory file structures, and ensuring data integrity in hierarchical databases.

🛠️ Constraints & Edge Cases

  • \(1 \le n \le 2000\)
  • \(0 \le \text{edges.length} \le 5000\)
  • Edge Cases: Single node tree (0 edges), cyclic graphs, disconnected components.

🧠 Approach & Intuition

The Aha! Moment

A graph with \(N\) nodes is a tree if and only if it has exactly \(N-1\) edges AND is fully connected.

🐢 Brute Force (Naive)

Try to find all cycles using DFS/BFS and check for reachability from a root. This is \(\mathcal{O}(N+E)\) but often implemented inefficiently by re-traversing components.

🐇 Optimal Approach

  1. If len(edges) != n - 1, return False immediately.
  2. Build an adjacency list.
  3. Perform a DFS or BFS from node 0.
  4. If the total number of unique visited nodes is \(n\), return True.

🧩 Visual Tracing

graph TD
    0 --- 1
    0 --- 2
    0 --- 3
    1 --- 4

💻 Solution Implementation

(Implementation details need to be added...)

⏱️ Complexity Analysis

  • Time Complexity: \(\mathcal{O}(V + E)\) where \(V\) is the number of nodes and \(E\) is the number of edges. We traverse all nodes and edges once.
  • Space Complexity: \(\mathcal{O}(V + E)\) for the adjacency list and recursion stack/visited set.

🎤 Interview Toolkit

  • Alternative: Disjoint Set Union (DSU). Union-Find is highly efficient here.
  • Edge Case Probe: A graph with 0 edges and 1 node is a valid tree.
  • [Number of Connected Components](../number_of_connected_components_in_graph/PROBLEM.md)
  • [Course Schedule II](../course_schedule_ii/PROBLEM.md)