計算進入該子樹能取到的最大的葉子是第幾大和最小的葉子是第幾大就行了
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
#define SF scanf
#define PF printf
using namespace std;
typedef long long LL;
const int MAXN = 200000;
struct Node {
int v, next;
} Edge[MAXN*2+10];
int adj[MAXN+10], ecnt;
int n, m;
void add(int u, int v) {
Node &e = Edge[++ecnt];
e.v = v; e.next = adj[u]; adj[u] = ecnt;
}
int dfs(int u, bool f) { // 0 第幾小 1 第幾大
if(!adj[u]) return 1;
int ret1 = 0, ret2 = n+1;
for(int i = adj[u]; i; i = Edge[i].next) {
int v = Edge[i].v;
if(f) ret1 += dfs(v, 0);
else ret2 = min(ret2, dfs(v, 1));
}
return f ? ret1 : ret2;
}
int main() {
SF("%d", &n);
for(int i = 1; i < n; i++) {
int u, v; SF("%d%d", &u, &v);
add(u, v);
}
for(int i = 1; i <= n; i++) if(!adj[i]) m++;
PF("%d %d\n", m+1-dfs(1, 0), dfs(1, 1));
}
/*
5
1 2
1 3
2 4
2 5
*/