开O2 40pts 剩余点WA ； 不开 20pts 剩余点RE

我的check就是从上到下每次删k个点，记录次数，这样不对么？

//#define LOCAL
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define mem(a, b) memset(a,b,sizeof(a))
#define sz(a) (int)a.size()
#define INF 0x3f3f3f3f
#define DNF 0x7f
#define DBG printf("this is a input\n")
#define fi first
#define se second
#define mk(a, b) make_pair(a,b)
#define pb push_back
#define LF putchar('\n')
#define SP putchar(' ')
#define p_queue priority_queue
#define CLOSE ios::sync_with_stdio(0); cin.tie(0)

template<typename T>
void read(T &x) {x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f *= -1;ch = getchar();}while(isdigit(ch)){x = x * 10 + ch - 48; ch = getchar();}x *= f;}
template<typename T, typename... Args>
void read(T &first, Args& ... args) {read(first);read(args...);}
template<typename T>
void write(T arg) {T x = arg;if(x < 0) {putchar('-'); x =- x;}if(x > 9) {write(x / 10);}putchar(x % 10 + '0');}
template<typename T, typename ... Ts>
void write(T arg, Ts ... args) {write(arg);if(sizeof...(args) != 0) {putchar(' ');write(args ...);}}
using namespace std;

const int N = 300005;
int n , high = 0;
int head[N] , cnt, dep[N];
int tot[N];
struct node
{
    int t , next;
}edge[N*2];
void add (int f, int t)
{
    edge[cnt].t = t;
    edge[cnt].next = head[f];
    head[f] = cnt ++;
}
void dfs (int u, int fa, int deep)
{
    dep[u] = deep;
    high = max (high , deep);
    for (int i = head[u] ; i != - 1; i = edge[i].next)
    {
        int v = edge[i].t;
        if(v != fa)
            dfs (v, u, deep + 1);
    }
}
int p[N];
bool check(int x)
{
    int len = high - 1;
    for (int i = 2 ; i <= high ; i ++)
        p[i-1] = tot[i];
    int index = 1, times = 0;
    while (index <= len)
    {
        ++ times;
        int sum = x;
        while (sum)
        {
            if (p[index] < sum)
                sum -= p[index], p[index] = 0 , index ++;
            else if(p[index] > sum)
                p[index] -= sum , sum = 0;
            else
                p[index] = 0, sum = 0, index ++;
        }
        if(index <= times)
            return false;
    }
    return times <= len;
}
int main()
{
    read (n);
    mem (head, -1);
    for (int i = 1 ; i < n ; i ++)
    {
        int u,  v;
        read (u , v);
        add (u , v);
        add (v , u);
    }
    dfs (1, 1, 1);
    for (int i = 1 ; i <= n ; i ++)
        tot[dep[i]]++;
    int l = 0, res = n , r = n;
    while(l<=r)
    {
        int mid = (l+r) / 2;
        if(check(mid))
        {
            res=mid;
            r=mid-1;
        }
        else l=mid+1;
    }
    write(res) , LF;
}
/*
2
6 5
0 1 0 1 0 1
1 3
3 5
1 2
2 4
4 5
 */