<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>

class Solution {
public:
    int dfsrec(int src, vector<vector<int>>& adj, vector<bool>& hasApple, vector<int>& mp, vector<bool>& visited){
        
        visited[src]=true;
        int total = 0;  // Initialize total time for the current subtree
    
    for (auto x : adj[src]) {
        if (!visited[x]) {
            int childTime = dfsrec(x, adj, hasApple, mp, visited);  // Recursively calculate time for the child node
            if (childTime > 0 || hasApple[x]) {
                total += childTime + 2;  // Accumulate time including the round trip to the child
            }
        }
    }

    mp[src] = total;  // Store the calculated time in mp for the current node

    // If the current node is not the root and it either has an apple or contributes to collecting apples, return the time
    if (src != 0 && (hasApple[src] || total > 0)) {
        return total;
    } else {
        return 0;
    }
        
    }
    int minTime(int n, vector<vector<int>>& edges, vector<bool>& hasApple) {
        vector<vector<int>> adj(n);
        vector<bool> visited(n,false);
        vector<int> mp(n,0);
        for(int i=0; i<edges.size(); i++){
            adj[edges[i][0]].push_back(edges[i][1]);
            adj[edges[i][1]].push_back(edges[i][0]);
        }
        
        
        visited[0]=true;
        dfsrec(0,adj,hasApple,mp,visited);
        
        return mp[0];
    }
};
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