【題目來自灰灰考研】

層次遍歷題目變形：

1.二叉樹採用二叉連結串列進行儲存（如下所示），每個結點包含資料域Data，左孩子指標域left和右孩子指標域right。請設計演算法判斷樹是否為完全二叉樹。

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)
{
	/*
		使用層次遍歷 
	*/
	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<<" ";
		/*
			如果當前節點為葉子節點，則佇列中剩餘的節點都是葉子節點
			如果當前節點只有左孩子，那麼這個節點一定是最後一個節點
			如果當前節點只有右孩子，那麼肯定不是完全二叉樹 
		*/ 
		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;
}