PAT 甲级 1110 Complete Binary Tree

1110 Complete Binary Tree （25 point(s)）

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

经验总结：

这一题。。。。真的不难，不论是用DFS还是BFS都可以，不过。。。自己傻了，想着刚做过的PAT 甲级 1102 Invert a Binary Tree 就用了scanf("%c %c"); 那一题题目中结点N<=10而且编号为0~9，可是这一题N<=20。。。。%c只能读取一位啊！于是自己写了半天，老是无法AC，还以为自己的思想出错了。。。最后才发现是这个错误，以后还是要再仔细一点！不能再犯这么低级的错误了！

AC代码

#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
int n,k;
struct node
{
	int lchild,rchild,no;
}tree[30];
bool flag[30]={false};
int BFS(int x)
{
	queue<node> q;
	q.push(tree[x]);
	int num=1;
	while(q.size())
	{
		node t=q.front();
		q.pop();
		if(t.lchild==-1&&t.rchild!=-1)
			return -1;
		else if(t.lchild==-1&&t.rchild==-1)
			if(num==n)
				return t.no;
			else
				return -1;
		else
		{
			++num;
			if(num==n)
				return t.lchild;
			q.push(tree[t.lchild]);
			if(t.rchild!=-1)
			{
				++num;
				if(num==n)
					return t.rchild;
				q.push(tree[t.rchild]);
			}
		}
	}
}
int main()
{
	char a[3],b[3];
	scanf("%d",&n);
	int t;
	for(int i=0;i<n;++i)
	{
		tree[i].no=i;
		scanf("%s %s",a,b);
		if(a[0]!='-')
		{
			sscanf(a,"%d",&t);
			flag[t]=true;
			tree[i].lchild=t;
		}
		else
			tree[i].lchild=-1;
		if(b[0]!='-')
		{
			sscanf(b,"%d",&t);
			flag[t]=true;
			tree[i].rchild=t;
		}
		else
			tree[i].rchild=-1;
	}
	int root;
	for(root=0;root<n;++root)
	{
		if(flag[root]==false)
			break;
	}
	int f=BFS(root);
	if(f!=-1)
		printf("YES %d\n",f);
	else
		printf("NO %d\n",root);
	return 0;
}