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:

9
7 8
- -
- -
- -
0 1
2 3
4 5
- -
- -

Sample Output 1:

YES 8

Sample Input 2:

8
- -
4 5
0 6
- -
2 3
- 7
- -
- -

Sample Output 2:

NO 1

#include<iostream>
#include<vector>
#include<string>
using namespace std;
vector<int> tree(21,-1), Left(20,-1), Right(20,-1), record(20,0);
int n;
void buildtree(int root){
if(Left[tree[root]]!=-1){
if(2*root+1>19)
return;
tree[2*root+1]=Left[tree[root]];
buildtree(2*root+1);
}
if(Right[tree[root]]!=-1){
if(2*root+2>19)
return;
tree[2*root+2]=Right[tree[root]];
buildtree(2*root+2);
}
}
int main(){
cin>>n;
string l, r;
for(int i=0; i<n; i++){
cin>>l>>r;
if(l[0]!='-'){
Left[i]=stoi(l);
record[Left[i]]=1;
}
if(r[0]!='-'){
Right[i]=stoi(r);
record[Right[i]]=1;
}
}
int i=0;
for(; i<n; i++)
if(record[i]==0)
break;
tree[0]=i;
buildtree(0);
int cnt=0;
for(int i=0; i<=20; i++)
if(tree[i]==-1)
if(i==n){
cout<<"YES "<<tree[n-1]<<endl;
return 0;
}
else{
cout<<"NO "<<tree[0]<<endl;
return 0;
}
}

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