RT，考虑记录每个进过的点的出度和，当前节点的深度为 $d$。那么我只要 出度和 $< d * k$ ($k$ 为二分的值即可)。

求大佬们 $hack$, 交上去只有 80pts

#include <bits/stdc++.h>
using namespace std;
#define N 300010
#define ll long long
#define int long long

template <class T>
inline void read(T& a){
	T x = 0, s = 1;
	char c = getchar();
	while(!isdigit(c)){ if(c == '-') s = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + (c ^ '0'); c = getchar(); }
	a = x * s;
	return ;
}

int n;
struct node{
	int v, next;
} t[N << 1];
int f[N];

int bian = 0;
inline void add(int u, int v){
	t[++bian] = (node){v, f[u]}, f[u] = bian;
	return ; 
}

int sum = 0;
bool flag = 0; 
int deth[N]; 
int du[N]; 

#define v t[i].v
void prepare(int now,  int father){
	deth[now] = deth[father] + 1;
	for(int i = f[now]; i; i = t[i].next){
		if(v != father){
			prepare(v, now);
			du[now]++; 
		}
	}
	return ; 
} 

bool dfs(int now, int father, int lim){
	if(deth[now] * lim >= n) return 1; 
	if(sum > deth[now] * lim) return 0; 
	sum += du[now]; 
	bool flag = 1;
	for(int i = f[now]; i; i = t[i].next){
		if(v != father)
			flag &= dfs(v, now, lim); 
	} 
	sum -= du[now]; 
	return flag; 
}
#undef v

bool check(int lim){
	sum = 0; 
	return dfs(1, 0, lim); 
}

signed main(){
//	freopen("hh.txt", "r", stdin); 
	read(n);
	for(int i = 1; i < n; i++){
		int x, y;
		read(x), read(y);
		add(x, y); add(y, x); 
	}
	prepare(1, 0); 
	int l = 0, r = n, ans = 0;
	while(l <= r){
		int mid = l + r >> 1;
		if(check(mid)) ans = mid, r = mid - 1;
		else l = mid + 1; 
	}
	cout << ans << endl;
	return 0; 
}