A1110 1110 Complete Binary Tree (25 point(s))

二叉树的建立，完全二叉树判断

1. 原文

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

2. 解析

• 查找根

  入度为0，indegree[i]

• 层序判断

   if (v[temp].lchild!=-1)
  {
  	q.push(v[temp].lchild);
  	if (flag==1)
  	{
  		complete=false;
  	}
  }else{
  	flag=1;
  }

3. AC代码

#include<cstdio>
#include<queue>
using namespace std;
char c1[5],c2[5];
struct node
{
	int lchild,rchild;
}v[30];
int getnum(char c[])
{
	int sum=0;
	for (int i = 0; c[i] != '\0'; ++i)
	{
		sum=sum*10+c[i]-'0';
	}
	return sum;
}
int indegree[30]={};
int main()
{
	int n;
	scanf("%d",&n);
	for (int i = 0; i < n; ++i)
	{
		scanf("%s%s",c1,c2);
		if (c1[0]=='-')
		{
			v[i].lchild=-1;
		}else{
			v[i].lchild=getnum(c1);
			indegree[getnum(c1)]++;
		}
		if (c2[0]=='-')
		{
			v[i].rchild=-1;
		}else{
			v[i].rchild=getnum(c2);
			indegree[getnum(c2)]++;
		}	
	}
	int root=0;
	for (int i = 0; i < n; ++i)
	{
		if (indegree[i]==0)
		{
			root=i;
			break;
		}
	}
	int flag=0;
	bool complete=true;
	queue<int> q;
	q.push(root);
	int last=0;
	while(!q.empty()){
		int temp=q.front();
		last=temp;
		q.pop();
		if (v[temp].lchild!=-1)
		{
			q.push(v[temp].lchild);
			if (flag==1)
			{
				complete=false;
			}
		}else{
			flag=1;
		}
		if (v[temp].rchild!=-1)
		{
			q.push(v[temp].rchild);
			if (flag==1)
			{
				complete=false;
			}
		}else{
			flag=1;
		}
	}
	if (complete)
	{
		printf("YES %d\n",last);
	}else{
		printf("NO %d\n",root);
	}

	return 0;
}