【PAT】【Advanced Level】1110. Complete Binary Tree (25)

1110. Complete Binary Tree (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Given a tree, you are supposed to tell if it is a complete binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=20) which is the total number of nodes in the tree -- and hence the nodes are numbered from 0 to N-1. Then N lines follow, each corresponds to a node, and gives the indices of the left and right children of the node. If the child does not exist, a "-" will be put at the position. Any pair of children are separated by a space.

Output Specification:

For each case, print in one line "YES" and the index of the last node if the tree is a complete binary tree, or "NO" and the index of the root if not. There must be exactly one space separating the word and the number.

Sample Input 1:

97 8- -- -- -0 12 34 5- -- -

Sample Output 1:

YES 8

Sample Input 2:

8- -4 50 6- -2 3- 7- -- -

Sample Output 2:

NO 1

原题链接：

https://www.patest.cn/contests/pat-a-practise/1110

思路：

树的遍历

满二叉树的特征

CODE：

#include<iostream>#include<cstdlib>#include<cstring>#include<string>#include<vector>#include<algorithm>#define N 22using namespace std;typedef struct S{int ind;int ls;int rs;int s;int fa;int tra;int fl;};S t[N];int tr;vector<int> ro;vector<int> le;vector<S> trav;void dfs1(int n,int flo){t[n].fl=flo;if (t[n].ls!=-1){dfs1(t[n].ls,flo+1);}t[n].tra=tr;tr++;trav.push_back(t[n]);if (t[n].rs!=-1){dfs1(t[n].rs,flo+1);}return ;}bool cmp(S a,S b){if(a.fl==b.fl){return a.tra<b.tra;}else{return a.fl<b.fl;}}int main(){int n;cin>>n;for (int i=0;i<n;i++) t[i].fa=-1;for (int i=0;i<n;i++){t[i].ind=i;t[i].s=0;string a,b;cin>>a>>b;if (a!="-"){t[i].ls=atoi(a.c_str());t[t[i].ls].fa=i;t[i].s++;}else{t[i].ls=-1;}if (b!="-"){t[i].rs=atoi(b.c_str());t[t[i].rs].fa=i;t[i].s++;}else{t[i].rs=-1;}//cout<<t[i].s<<endl;if (t[i].s!=0){ro.push_back(i);}else{le.push_back(i);}}int root;for (int i=0;i<n;i++){if (t[i].fa==-1){root=i;break;}}tr=0;dfs1(root,0);sort(trav.begin(),trav.end(),cmp);//cout<<trav[n-1].ind<<endl;int flag=0;for (int i=0;i<ro.size();i++){if (t[ro[i]].s!=2){if (!(t[ro[i]].ls==trav[n-1].ind)){//cout<<ro[i]<<" "<<t[ro[i]].ls<<" "<<t[ro[i]].rs<<endl;flag=1;break;}}}if (flag==0){for (int i=0;i<le.size();i++){if (t[le[i]].fl<(trav[n-1].fl-1)){flag=1;break;}else if (t[le[i]].fl==(trav[n-1].fl-1)){if (t[le[i]].tra<trav[n-1].tra){flag=1;break;}}}}if (flag==0){cout<<"YES "<<trav[n-1].ind<<endl;}else{cout<<"NO "<<root<<endl;}return 0;}