題解：

根據題目，我們可以知道只要求不是根的點所連點子鏈個數-1再加上根所連的子鏈個數-2為刪條的條數，就是可以這個樹可以分成多少條鏈，所刪條數*2+1就是答案了。

#include <algorithm>
#include  <iostream>
#include   <cstdlib>
#include   <cstring>
#include    <cstdio>
#include    <string>
#include    <vector>
#include    <bitset>
#include     <stack>
#include     <cmath>
#include     <deque>
#include     <queue>
#include      <list>
#include       <set>
#include       <map>
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define line printf("---------------------------\n")
#define mem(a, b) memset(a, b, sizeof(a))
#define pi acos(-1)
using namespace std;
typedef long long ll;
const double eps = 1e-9;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 1000000 + 10;

int du[maxn], head[maxn], ans, s, cnt;
struct Edge{
	int v, next;
	Edge(){};
	Edge(int v, int next):v(v), next(next){};
}edge[maxn*2];

int dfs(int u, int father) {
	int sum = 0;
	for(int i = head[u]; ~i; i = edge[i].next){
		int v = edge[i].v;
		if(v != father){
			sum += dfs(v, u);
		}
	}
	if(sum >= 2){
		if(s == u){
			ans += sum-2;
		}else{
			ans += sum-1;
		}
		return 0;
	}else{
		return 1;
	}
}
void init(int n){
	ans = cnt = 0;
	for(int i = 0; i <= n; i++){
		head[i] = -1;
		du[i] = 0;
	}
}
void InsertEdge(int u, int v){
	edge[cnt] = Edge(v, head[u]);
	head[u] = cnt++;
}

int main(){
	int t;
	scanf("%d", &t);
	while(t--){
		int n;
		scanf("%d", &n);
		init(n);
		for(int i = 0; i < n-1; i++){
			int u, v;
			scanf("%d %d", &u, &v);
			du[u]++;
			du[v]++;
			InsertEdge(u, v);
			InsertEdge(v, u);
		}
		for(int i = 1; i <= n; i++){
			if(du[i] == 1){
				s = i;
				break;
			}
		}
		dfs(s, -1);
		printf("%d\n", ans*2+1);
	}
}