【链接】 我是链接,点我呀:)
【题意】

【题解】

找到度数为1的点。
他们显然是叶子节点。
然后每个叶子节点。
往上进行bfs.
累计他们的父亲节点的儿子的个数。
如果都满足要求那么就继续往上走。
直到不能走。或已经走了k步。
且要求走了k步之后。他们都到了同一个节点。(根节点

这道题。
n=1的时候,认为是无解的。
(因为题目中说"some vertices of degree 1",也就是说必须>= 1.......

【代码】

/*find each vertex i where du[i]==1bfs(i) one depth and get y,cnt[y]++;after thatcount the number if different cnt[y] ->CNT (only consider cnt[y]>0 if (CNT!=1) return 0;set the only cnt value as kkwe can now know that the root has kk childrenfind the root,(du[root]==kk)(if there are more than one i such that du[i]==kk,no solution(if there are no such i,such that du[i]==k,no solutiondfs from root.if (dep!=k && du[x]==1) return 0;//must have k depthcheck if every vertex except the bottom vertext all has kk children*/
#include <bits/stdc++.h>
#define ll long long
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
using namespace std;const int N = 1e5;int n,k;
vector<int> g[N+10];
bool flag[N+10];
vector<int> v,tempv;
int cnt[N+10];void wujie(){puts("No");exit(0);
}int main(){#ifdef ccyfreopen("rush.txt","r",stdin);#endifscanf("%d%d",&n,&k);if (n==1){puts("No");return 0;}rep1(i,1,n-1){int x,y;scanf("%d%d",&x,&y);g[x].push_back(y);g[y].push_back(x);}rep1(i,1,n)if ((int)g[i].size()==1){v.push_back(i);}rep1(dep,1,k){/*for (int x:v){printf("%d ",x);}puts("");*/rep1(j,0,(int)v.size()-1){int x = v[j];int cntup = 0;rep1(l,0,(int)g[x].size()-1){int y = g[x][l];
//              printf("%d ",y);if (flag[y]) continue;cnt[y]++;cntup++;}       if (cntup!=1) wujie();flag[x] = true;}
//      puts("");v.clear();rep1(i,1,n){if (cnt[i]!=0 && cnt[i]<3) wujie();if (cnt[i]==0) continue;v.push_back(i);cnt[i] = 0;                     }}rep1(i,0,(int)v.size()-1){if (v[i]!=v[0]) wujie();}puts("Yes");return 0;
}