<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:
    unordered_map<int, vector<int>> g; // to store the graph
    unordered_map<int, bool> visited; // to stop exploring same nodes again and again.
	
    void createGraph(vector<vector<int>>& edges) {
      for (auto e: edges) {
        g[e[0]].push_back(e[1]); // adjecency list representation
		g[e[1]].push_back(e[0]); // adjecency list representation
      }
    }
  
    int dfs(int node, int myCost, vector<bool>& hasApple) {
	  if (visited[node]) {
		  return 0;
	  }
	  visited[node] = true;
	  
      int childrenCost = 0; // cost of traversing all children. 
      for (auto x: g[node]) { 
        childrenCost += dfs(x, 2, hasApple);  // check recursively for all apples.
      }

      if (childrenCost == 0 && hasApple[node] == false) {
	  // If no child has apples, then we won't traverse the subtree, so cost will be zero.
	  // similarly, if current node also does not have the apple, we won't traverse this branch at all, so cost will be zero.
        return 0;
      }
	  
	  // Children has at least one apple or the current node has an apple, so add those costs.
      return (childrenCost + myCost);
    }
  
    int minTime(int n, vector<vector<int>>& edges, vector<bool>& hasApple) {
      createGraph(edges); // construct the graph first.
      return dfs(0, 0, hasApple); // cost of reaching the root is 0. For all others, its 2.
    }
};
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