1110. Complete Binary Tree (25)

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

---

给出一棵树，判断是否是完全二叉树。用BFS的方法遍历这棵树，计算遍历过的节点数，同时更新遍历的最后一个节点的值，遇到某点为-1时跳出循环。最后如果遍历过的点等于总结点数，说明是完全二叉树，否则不是。

代码：

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <queue>
using namespace std;

struct node
{
	int val;
	int left,right;
	node():left(-1),right(-1){}
};

int main()
{
	int n;
	cin>>n;
	vector<node>tree(n);
	vector<bool>isroot(n,true);
	for(int i=0;i<n;i++)
	{
		string l,r;
		cin>>l>>r;
		if(l[0]!='-')
		{
			int left=atoi(l.c_str());
			tree[i].left=left;	
			isroot[left]=false;
		}
		if(r[0]!='-')
		{
			int right=atoi(r.c_str());
			tree[i].right=right;
			isroot[right]=false;
		}
	}
	int root;
	for(int i=0;i<n;i++)
	{
		if(isroot[i])
		{
			root=i;
			break;
		}
	}
	queue<int>que;
	que.push(root);
	int count=0,last;
	while(!que.empty())
	{
		int tmp=que.front();
		if(tmp==-1)
		{
			break;
		}
		last=tmp;
		count++;
		que.pop();
		que.push(tree[tmp].left);
		que.push(tree[tmp].right);
	}
	if(count==n)
	{
		cout<<"YES "<<last;
	}
	else
	{
		cout<<"NO "<<root;
	}
}