Naive Approach

A brute-force approach would involve exploring all possible paths from vertex 0, collecting apples along the way, and returning to vertex 0. This approach would be highly inefficient, especially for larger trees, as it would involve redundant calculations and exploring paths that don't necessarily lead to any apples.

Optimal Approach

The optimal approach utilizes Depth-First Search (DFS) to traverse the tree. The key idea is to recursively explore the tree, and for each subtree, determine if it's worth visiting (i.e., if it contains any apples, either directly or in its subtrees). If a subtree contains an apple, the cost of traversing that subtree (twice the edge weight to enter and exit) is added to the total time. The algorithm starts at vertex 0 and explores the tree.

Algorithm

1. Build an adjacency list representation of the tree from the given edges.
2. Define a recursive DFS function that takes the current node and its parent as input.
3. In the DFS function:
   • Initialize a subtree_cost to 0.
   • Iterate through the neighbors of the current node (excluding the parent).
   • Recursively call the DFS function on each neighbor.
   • Add the returned cost from the recursive call to the subtree_cost.
   • If the current node has an apple or the subtree_cost is greater than 0 (meaning there are apples in the subtree), add 2 to the subtree_cost (representing the cost of entering and exiting the current node's edge).
4. Return the subtree_cost.
5. Call the DFS function starting from vertex 0 with a parent of -1.
6. If the result is > 0, then return it, otherwise return 0.

Edge Cases

• Empty Tree: If the tree is empty (n=0), return 0.
• No Apples: If no vertices have apples, return 0.
• Single Node Tree: If the tree has only one node (n=1) and it doesn't have an apple, return 0. If it has an apple, return 0 (as we start and end at the same node).

Code Example (Python)

def minTime(n: int, edges: list[list[int]], hasApple: list[bool]) -> int:
    adj = [[] for _ in range(n)]
    for u, v in edges:
        adj[u].append(v)
        adj[v].append(u)

    def dfs(node: int, parent: int) -> int:
        subtree_cost = 0
        for neighbor in adj[node]:
            if neighbor != parent:
                cost = dfs(neighbor, node)
                subtree_cost += cost

        if (hasApple[node] or subtree_cost > 0) and parent != -1:
            subtree_cost += 2

        return subtree_cost

    result = dfs(0, -1)
    return max(0, result)

Big O Analysis

• Time Complexity: O(N), where N is the number of vertices in the tree. The DFS visits each node once.
• Space Complexity: O(N). The space complexity is dominated by the adjacency list, which takes O(N) space, and the recursion stack, which can take up to O(N) space in the worst-case scenario (a skewed tree).