<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>

#define pb push_back
#define ll long long
#define f(i,s,e) for(long long i=s;i<e;i++)

class Solution {
    vector<vector<int>> adj;
    int n;
    vector<int> dp;  // dp vector to store the minimum time at each node

    // DFS function to calculate the minimum time
    void dfs(int root, int par, vector<bool>& has) {
        // Traverse the adjacent nodes (children)
        for (auto it : adj[root]) {
            if (it != par) dfs(it, root, has);
        }
        int cnt = 0;
        // After traversing all children, calculate the dp value for the current node
        for (auto it : adj[root]) {
            if (it != par && dp[it] != -1) {
                    cnt += dp[it] + 2;  // Accumulate time with the current child
            }
        }
        if(cnt != 0)dp[root] = cnt;
        // If this node has an apple and dp[root] is 0, mark it with 1
        if (dp[root] == -1 && has[root]) dp[root] = 0;
    }

public:
    int minTime(int n, vector<vector<int>>& edges, vector<bool>& hasApple) {
        this->n = n;
        adj.resize(n);
        dp.resize(n, -1);  // Initialize dp vector with size n and all values 0

        // Construct the adjacency list from the edges
        for (auto it : edges) {
            adj[it[0]].pb(it[1]);
            adj[it[1]].pb(it[0]);
        }

        // Call the DFS function starting from node 0 and parent -1
        dfs(0, -1, hasApple);

        // Return the total time at the root (node 0)
        return max(0,dp[0]);
    }
};

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