tree.cpp

里面装有 5 个小程序

// Tree.cpp : Defines the entry point for the console application.
//

/*
  实验题目：二叉的遍历
  开发思想：本实验分别用了递归和非递归来实现前序、中序、后序
			方式的遍历二叉树，还实现了层次遍历，并求出树高。
			在这些实现中，主要用到的是队列链和链栈。
  开发人员：葛兴高
  开发日期：2004、10、29
*/
#include "stdafx.h"


void main()
{
	BuildTree B;
	int Hight;
	char y;
	int flag=1;
	int choice=1;
	cout<<"\t\t为了确保程序可以正常进行，确认你输入的二叉树是正确的!\n";
	cout<<"\t\t下面给出几个例子，可以参照输入：\n";

	cout<<"\t\t\t ABD..E..C.F..#\n";
	cout<<"\t\t\t ABDF..G..E..C#\n";
	cout<<"\t\t\t ABD..EG..H..CF...#\n\n";
	cout<<"\t\t   *******注意要以 # 结束*******\n";
	cout<<"\n\t\t你想要进入程序吗？(y/n):";
	cin>>y;
	cout<<"\n\n\n\n\n\n\n\n\n\n\n\n\n\n";
	if(y=='y'||y=='Y')
	{
	while(choice!=0)
	{
		if(flag==1){
		cout<<"\t\t\t1 ...建立二叉树\n";
		cout<<"\t\t\t2 ...递归前序遍历\n";
		cout<<"\t\t\t3 ...非递归前序遍历\n";
		cout<<"\t\t\t4 ...递归中序遍历\n";
		cout<<"\t\t\t5 ...非递归中序遍历\n";
		cout<<"\t\t\t6 ...递归后序遍历\n";
		cout<<"\t\t\t7 ...非递归后序遍历\n";
		cout<<"\t\t\t8 ...层次遍历\n";
		cout<<"\t\t\t9 ...求树高\n";
		cout<<"\t\t\t0...退出程序\n";
		}
		cout<<"\t\t\t请输入你的选择(0-9)：";
		cin>>choice;
		
		switch(choice)
		{
			
		case 1:
			B.Create();
			flag=0;
			break;
		case 2:
			if(flag==0)
			{
				cout<<"递归前序遍历：\n";
				B.PreOrder_1(B.head,1,0);
				cout<<endl;
			//	getch();
			}
			else
				cout<<"\n\t\t\t********请先建二叉树!********\n\n";
			break;
		case 3:
			if(flag==0)
			{
				cout<<endl<<"非递归前序遍历：\n";
				B.PreOrder_2(B.head);
				cout<<endl;
			//	getch();
			}
			else
				cout<<"\n\t\t\t********请先建二叉树!********\n\n";
			break;
		case 4:
			if(flag==0)
			{
				cout<<endl<<"递归中序遍历:\n"<<endl;
				B.InOrder_1(B.head,1,0);
				cout<<endl;
			//	getch();
			}
			else
				cout<<"\n\t\t\t********请先建二叉树!********\n\n";
			break;
		case 5:
			if(flag==0)
			{
				cout<<endl<<"非递归中序遍历:\n"<<endl;
				B.InOrder_2(B.head);
				cout<<endl;
			//	getch();
			}
			else
				cout<<"\n\t\t\t********请先建二叉树!********\n\n";
			break;
		case 6:
			if(flag==0)
			{
				cout<<endl<<"递归后序遍历:\n";
				B.PostOrder_1(B.head,1,0);
				cout<<endl;
			//	getch();
			}
			else
				cout<<"\n\t\t\t********请先建二叉树!********\n\n";
			break;
		case 7:
			if(flag==0)
			{
				cout<<endl<<"非递归后序遍历：\n";
				B.PostOrder_1(B.head,1,0);
				cout<<endl;
			//	getch();
			}
			else
				cout<<"\n\t\t\t********请先建二叉树!********\n\n";
			break;
		case 8:
			if(flag==0)
			{
				cout<<endl<<"层次遍历：\n";
				B.LayerOrder();
				cout<<endl;
			//	getch();
			}
			else
				cout<<"\n\t\t\t********请先建二叉树!********\n\n";
			break;
		case 9:
			if(flag==0)
			{
				cout<<endl<<"树高：\n";
				Hight=B.GetHigh(B.head,1);
				cout<<Hight;
				cout<<endl;
			//	getch();
			}
			else
				cout<<"\n\t\t\t********请先建二叉树!********\n\n";
			break;
		case 0:
			break;
		default:
			choice=1;
		}
	}
	}
}