<p>Given an undirected tree consisting of <code>n</code> vertices numbered from <code>0</code> to <code>n-1</code>, which has some apples in their vertices. You spend 1 second to walk over one edge of the tree. <em>Return the minimum time in seconds you have to spend to collect all apples in the tree, starting at <strong>vertex 0</strong> and coming back to this vertex.</em></p>

<p>The edges of the undirected tree are given in the array <code>edges</code>, where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> means that exists an edge connecting the vertices <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code>. Additionally, there is a boolean array <code>hasApple</code>, where <code>hasApple[i] = true</code> means that vertex <code>i</code> has an apple; otherwise, it does not have any apple.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/04/23/min_time_collect_apple_1.png" style="width: 300px; height: 212px;" />
<pre>
<strong>Input:</strong> n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,true,false,true,true,false]
<strong>Output:</strong> 8 
<strong>Explanation:</strong> The figure above represents the given tree where red vertices have an apple. One optimal path to collect all apples is shown by the green arrows.  
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/04/23/min_time_collect_apple_2.png" style="width: 300px; height: 212px;" />
<pre>
<strong>Input:</strong> n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,true,false,false,true,false]
<strong>Output:</strong> 6
<strong>Explanation:</strong> The figure above represents the given tree where red vertices have an apple. One optimal path to collect all apples is shown by the green arrows.  
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,false,false,false,false,false]
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>edges.length == n - 1</code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub> &lt; b<sub>i</sub> &lt;= n - 1</code></li>
	<li><code>hasApple.length == n</code></li>
</ul>

'''
Input: n: int, edges: List[List[int]], List[bool], 
Output: min time to take all apples and get back: int

clarify: 
- n range? 1 to 10^5
- vals 0 to n-1?
- length of edges?
- length of hasApple == n?

examples:
n = 1, edges = [[]], hasApple = [] -> 0

n = 2, edges = [[0,1]], hasApple = [false,true] -> 2

approaches:
1. dfs
n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]] -> 6
hasApple = [false,false,true,false,false,true,false]
Each edge takes 2 seconds -> we need to count edges to apples
                0
            1      (2)
          4  (5)   3   6
1. create a adj map of {node:[children]}
adj = {0:[1,2],1:[0,4,5],2:[0,3,6],3:[2],4:[1],5:[1],6:[2]}

2. dfs while checking visited set
dfs(node) -> int

'''
class Solution:
    def minTime(self, n: int, edges: List[List[int]], hasApple: List[bool]) -> int:
        adj = defaultdict(list)
        for u, v in edges:
            adj[u].append(v)
            adj[v].append(u)

        visited = set()

        def dfs(node):
            visited.add(node)
            secs = 0
            for child in adj[node]:  # Use get() for safe access
                if child not in visited:
                    secs += dfs(child)
            if secs > 0:
                return secs + 2
            return 2 if hasApple[node] else 0

        return max(dfs(0) - 2, 0)
57455