Graph Valid Tree

 

1. Clarify the problem: Before diving into solving the problem, let's clarify the requirements:

  • We are given an undirected graph in the form of an adjacency list and the number of nodes in the graph.
  • Our task is to determine if the given graph is a valid tree.
  • A valid tree must satisfy two conditions:
    1. The graph must be connected, i.e., there is a path between any two nodes.
    2. The graph must not contain any cycles.

2. Analyze the problem: To solve this problem, we can use a depth-first search (DFS) or breadth-first search (BFS) algorithm to traverse the graph and check for cycles and connectivity.

  • We can start the search from any node in the graph.
  • If all nodes are visited during the traversal and there are no cycles, the graph is a valid tree.

3. Design an algorithm: Here is the algorithm to solve the problem:

  1. Create a set to keep track of visited nodes.
  2. Perform a DFS or BFS traversal starting from any node in the graph.
  3. During the traversal, mark each visited node as visited.
  4. If a node is visited again during the traversal, there is a cycle in the graph, and it is not a valid tree.
  5. After the traversal, check if all nodes have been visited. If not, the graph is not connected, and it is not a valid tree.
  6. If all nodes have been visited and no cycles are found, the graph is a valid tree.

4. Explain your approach: The approach involves performing a DFS or BFS traversal of the graph, starting from any node. We mark each visited node as visited during the traversal. If a node is visited again, there is a cycle in the graph, and it is not a valid tree. After the traversal, we check if all nodes have been visited. If not, the graph is not connected, and it is not a valid tree. If all nodes have been visited and no cycles are found, the graph is a valid tree.

5. Write clean and readable code:

python
def isValidTree(n, edges):
    # Step 1: Create adjacency list representation of the graph
    adjacency = {i: [] for i in range(n)}
    for edge in edges:
        u, v = edge
        adjacency[u].append(v)
        adjacency[v].append(u)

    # Step 2: Perform DFS or BFS traversal
    visited = set()

    def dfs(node, parent):
        visited.add(node)
        for neighbor in adjacency[node]:
            if neighbor != parent:
                if neighbor in visited:
                    return False
                if not dfs(neighbor, node):
                    return False
        return True

    if not dfs(0, -1):
        return False

    # Step 3: Check if all nodes are visited
    return len(visited) == n

6. Test your code: Let's test the code with different test cases:

  1. n = 5, edges = [[0,1],[0,2],[0,3],[1,4]]
    • Expected output: True
    • Explanation: The graph is a valid tree as it is connected, and there are no cycles.
  2. n = 5, edges = [[0,1],[1,2],[2,3],[1,3],[1,4]]
    • Expected output: False
    • Explanation: The graph is not a valid tree as there is a cycle present (1 -> 2 -> 3 -> 1).
  3. n = 1, edges = []
    • Expected output: True
    • Explanation: The graph is a valid tree with a single node.
  4. n = 3, edges = [[0,1],[1,2]]
    • Expected output: False
    • Explanation: The graph is not a valid tree as it is not connected (node 2 is not reachable from node 0).

7. Optimize if necessary: The initial solution already follows an optimal approach using DFS traversal, so there is no further optimization required.

8. Handle error cases: The code handles the case where the input edges list is empty, representing a graph with no edges. In this case, it returns True since a graph with no edges is considered a valid tree.

9. Discuss complexity analysis:

  • The time complexity of the solution is O(V + E), where V is the number of nodes (vertices) and E is the number of edges in the graph. We perform a DFS traversal, which visits each node and edge once.
  • The space complexity is O(V), as we use a set to keep track of visited nodes and the adjacency list to represent the graph.
  • The solution has an optimal time and space complexity.
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