数据结构-判断是否为完全二叉树

【题目来自灰灰考研】

层次遍历题目变形：

1.二叉树采用二叉链表进行存储（如下所示），每个结点包含数据域Data，左孩子指针域left和右孩子指针域right。请设计算法判断树是否为完全二叉树。

Typedef struct BitNode{   

TElemType data;  

      struct BitNode *left, *right;

} *BiTree ;

#include<iostream>
#include<cstdio>
#include<cstdlib>
#define MAXSIZE 20
using namespace std;

typedef struct TNode{
	char data;
	struct TNode *lChild, *rChild;
}TNode;

TNode *createBTree()
{
	//124##57###38#9###
	TNode *node;
	char data;
	cin>>data;
	if(data == '#')
		return NULL;
	else
	{
		node = (TNode*)malloc(sizeof(TNode));
		node->data = data;
		node->lChild = createBTree();
		node->rChild = createBTree();
	}
	return node;
}


bool isCompleteBTree(TNode *root)
{
	/*
		使用层次遍历 
	*/
	int front = 0, rear = 0;
	TNode *queue[MAXSIZE], *p;
	rear = (rear + 1) % MAXSIZE;
	queue[rear] = root;
	while(rear != front)
	{
		front = (front + 1) % MAXSIZE;
		p = queue[front];
//		cout<<p->data<<" ";
		/*
			如果当前节点为叶子节点，则队列中剩余的节点都是叶子节点
			如果当前节点只有左孩子，那么这个节点一定是最后一个节点
			如果当前节点只有右孩子，那么肯定不是完全二叉树 
		*/ 
		if(p->lChild && p->rChild)
		{
			rear = (rear + 1) % MAXSIZE;
			queue[rear] = p->lChild;
			rear = (rear + 1) % MAXSIZE;
			queue[rear] = p->rChild;
		}
		else if(p->lChild && p->rChild == NULL)
		{
			int count = (rear - front + MAXSIZE) % MAXSIZE;
			if(count)
				return false;
		}
		else if(p->rChild && p->lChild == NULL)
		{
			return false;
		}
		else if(p->rChild == NULL && p->lChild == NULL)
		{
			while(rear != front)
			{
				front = (front + 1) % MAXSIZE;
				p = queue[front];
				if(p->rChild || p->rChild)
					return false;
			}
		}
	}
	return true;
}

int main()
{
	TNode *root;
	root = createBTree();
	bool result = isCompleteBTree(root);
	cout<<"If is A Complete Binary Tree?："<<boolalpha<<result<<endl;
	return 0;
}