You are given a tree (a graph with n vertices and n - 1 edges in which it's possible to reach any vertex from any other vertex using only its edges).
A vertex can be destroyed if this vertex has even degree. If you destroy a vertex, all edges connected to it are also deleted.
Destroy all vertices in the given tree or determine that it is impossible.
The first line contains integer n (1 ≤ n ≤ 2·105) — number of vertices in a tree.
The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ n). If pi ≠ 0 there is an edge between vertices i and pi. It is guaranteed that the given graph is a tree.
If it's possible to destroy all vertices, print "YES" (without quotes), otherwise print "NO" (without quotes).
If it's possible to destroy all vertices, in the next n lines print the indices of the vertices in order you destroy them. If there are multiple correct answers, print any.
5 0 1 2 1 2
YES 1 2 3 5 4
4 0 1 2 3
NO
In the first example at first you have to remove the vertex with index 1 (after that, the edges (1, 2) and (1, 4) are removed), then the vertex with index 2 (and edges (2, 3) and (2, 5) are removed). After that there are no edges in the tree, so you can remove remaining vertices in any order.
题解: 以其中一个点为根,从叶子节点开始遍历。若此节点度数为奇数, 则不操作,若是偶数,则一定要删除,否则,若删除他的父亲节点 则此点变为奇数就不能删掉了。 代码: #include<bits/stdc++.h> using namespace std; const int maxn=200005; int u[maxn],par[maxn]; bool vis[maxn]; vector<int>rode[maxn]; stack<int>P;//存储遍历顺序 queue<int>Q; void dfs(int v,int pre) { P.push(v);par[v]=pre; for(int i=0;i<rode[v].size();i++) { if(rode[v][i]==pre)continue; dfs(rode[v][i],v); } } void dfs2(int x) { Q.push(x);vis[x]=1; for(int i=0;i<rode[x].size();i++) { int to=rode[x][i]; u[to]--; if(to==par[x]||vis[to])continue; if(u[to]%2==0) dfs2(to); } } int main() { int n;scanf("%d",&n); for(int i=1;i<=n;i++) { int x;scanf("%d",&x); if(x) { rode[i].push_back(x),rode[x].push_back(i); u[i]++,u[x]++; } } dfs(1,0); while(!P.empty()) { int x=P.top();P.pop(); //printf("%d\n",x); if(u[x]%2==0) dfs2(x); } //printf("%d\n",Q.size()); if(Q.size()==n) { printf("YES\n"); while(!Q.empty()) { printf("%d\n",Q.front()); Q.pop(); } } else printf("NO\n"); return 0; }