Given an undirected tree consisting of n vertices numbered from 0 to n-1, which has some apples in their vertices. You spend 1 second to walk over one edge of the tree. Return the minimum time in seconds you have to spend to collect all apples in the tree, starting at vertex 0 and coming back to this vertex.
The edges of the undirected tree are given in the array edges, where edges[i] = [ai, bi] means that exists an edge connecting the vertices ai and bi. Additionally, there is a boolean array hasApple, where hasApple[i] = true means that vertex i has an apple; otherwise, it does not have any apple.
Example 1:
Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,true,false,true,true,false]
Output: 8
Explanation: 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.
Example 2:
Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,true,false,false,true,false]
Output: 6
Explanation: 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.
class Solution
{
public:
int dfs(vector<int> v[], int node, vector<bool> hasApple, vector<bool> &vs)
{
vs[node] = true;
int x = 0;
for (int i = 0; i < v[node].size(); i++)
{
if (!vs[v[node][i]])
{
x += dfs(v, v[node][i], hasApple, vs);
}
}
if (hasApple[node] || x != 0)
{
x = x + 2;
}
return x;
}
int minTime(int n, vector<vector<int>> &edges, vector<bool> &hasApple)
{
vector<int> v[n];
vector<bool> vs(n);
for (int i = 0; i < edges.size(); i++)
{
v[edges[i][0]].push_back(edges[i][1]);
v[edges[i][1]].push_back(edges[i][0]);
}
int sum = dfs(v, 0, hasApple, vs);
if (sum != 0)
{
return sum - 2;
}
return sum;
}
};
