将一棵树变成一个环

A tree with N nodes and N-1 edges is given. To connect or disconnect one edge, we need 1 unit of cost respectively. The nodes are labeled from 1 to N. Your job is to transform the tree to a cycle(without superfluous edges) using minimal cost.

A cycle of n nodes is defined as follows: (1)a graph with n nodes and n edges (2)the degree of every node is 2 (3) each node can reach every other node with these N edges.

InputThe first line contains the number of test cases T( T<=10 ). Following lines are the scenarios of each test case.
In the first line of each test case, there is a single integer N( 3<=N<=1000000 ) - the number of nodes in the tree. The following N-1 lines describe the N-1 edges of the tree. Each line has a pair of integer U, V ( 1<=U,V<=N ), describing a bidirectional edge (U, V).
OutputFor each test case, please output one integer representing minimal cost to transform the tree to a cycle.
Sample Input

1
4
1 2
2 3
2 4

Sample Output

3

        
 

Hint

In the sample above, you can disconnect (2,4) and then connect (1, 4) and
(3, 4), and the total cost is 3.

题意 ： 给你一颗树，要求将此树变成一个环，并且要求每个点的度为2，删去或增加一条边的花费都是1，求最小的花费。

思路分析：思维题，对于树上的点如果其不是根节点并且具有3个分支，则其一定要去掉一个，保留两个，然后其在返回父节点的时候它是不给父节点在提供任何分支的，否则会提供一个分支。

代码示例：

const int maxn = 1e6+5;

int n;
vector<int>ve[maxn];
int ans;

int dfs(int x, int fa){
    int sum = 0;
    
    for(int i = 0; i < ve[x].size(); i++){
        int to = ve[x][i];
        if (to == fa) continue; 
        sum += dfs(to, x);
    }
    if (x != 1 && sum >= 2){
        ans += sum-1;
        return 0;        
    }
    else if (x == 1 && sum > 2){
        ans += sum-2;
    }
    return 1; 
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int t;
    int x, y;
    
    cin >> t;
    while(t--){
        scanf("%d", &n);
        for(int i = 1; i <= n; i++) ve[i].clear();
        for(int i = 1; i < n; i++){
            scanf("%d%d", &x, &y);
            ve[x].push_back(y);
            ve[y].push_back(x);        
        }  
        ans = 0;
        dfs(1, 0);
        printf("%d
", ans*2+1);
    }
    return 0;
}