pat甲1110. Complete Binary Tree（完全二叉树）

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:

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

tips:这道题用到了堆的思想，左儿子的为父亲的二倍，右儿子为父亲的二倍+1；当出现一个节点的>n时，则不是完全二叉树，最为为n的点一定是最后一个结点

#include<iostream>
#include<cstring>
#include<string>

using namespace std;
const int maxn=200;
int n,rt;
int t[maxn][2];
int book[maxn];

int flag=1;

void dfs(int x,int cnt)
{
if(!flag)return;
if(cnt>n){
flag=0;
return;
}
if(cnt==n)flag=x;
if(t[x][0]!=-1)dfs(t[x][0],cnt*2);
if(t[x][1]!=-1)dfs(t[x][1],cnt<<1|1);
}
int main()
{
cin>>n;
for(int i=0;i<n;i++)
{
char a[4],aa[4];
scanf("%s %s",a,aa);
if(a[0]=='-')t[i][0]=-1;
else  sscanf(a,"%d",&t[i][0]);

if(aa[0]=='-')t[i][1]=-1;
else  sscanf(aa,"%d",&t[i][1]);

book[t[i][0]]=book[t[i][1]]=1;
}
for(int i=0;i<n;i++)if(!book[i]){
rt=i;break;
}
dfs(rt,1);
if(!flag)cout<<"NO"<<" "<<rt<<endl;
else cout<<"YES"<<" "<<flag<<endl;
return 0;
}