#include<string.h>
 #include<ctype.h>
 #include<malloc.h> 
 #include<limits.h> 
 #include<stdio.h> 
 #include<stdlib.h> 
 #include<io.h> 
 #include<math.h>
 #include<process.h> 
 #include<iostream.h>
 // 函數結果狀態代碼
 #define TRUE 1
 #define FALSE 0
 #define OK 1
 #define ERROR 0
 #define INFEASIBLE -1

 typedef int Status; // Status是函數的類型,其值是函數結果狀態代碼，如OK等
 typedef int Boolean;
 #define CHAR 0 // 整型(二者選一)
 #if CHAR
   typedef char TElemType;
   TElemType Nil=' '; // 設字符型以空格符為空
 #else
   typedef int TElemType;
   TElemType Nil=0; // 設整型以0為空
 #endif
 #define MAX_TREE_SIZE 100 // 二叉樹的最大結點數
 typedef TElemType SqBiTree[MAX_TREE_SIZE]; // 0號單元存儲根結點

 struct position
 {
   int level,order; // 結點的層,本層序號(按滿二叉樹計算)
 };
 #include"樹.cpp"

 Status visit(TElemType e)
 {
   cout<<e<<' ';
   return OK;
 }

 void main()
 {
   Status i;
   int j;
   position p;
   TElemType e;
   SqBiTree T,s;
   InitBiTree(T);
   CreateBiTree(T);
   cout<<"建立二叉樹后,樹空否？"<<BiTreeEmpty(T)<<"(1:是 0:否) 樹的深度="<<BiTreeDepth(T)<<endl;
   i=Root(T,e);
   if(i)
     cout<<"二叉樹的根為："<<e<<endl;
   else
     cout<<"樹空，無根"<<endl;
   cout<<"層序遍歷二叉樹:"<<endl;
   LevelOrderTraverse(T,visit);
   cout<<"中序遍歷二叉樹:"<<endl;
   InOrderTraverse(T,visit);
   cout<<"后序遍歷二叉樹:"<<endl;
   PostOrderTraverse(T,visit);
   cout<<"請輸入待修改結點的層號 本層序號: ";
   cin>>p.level>>p.order;
   e=Value(T,p);
   cout<<"待修改結點的原值為"<<e<<"請輸入新值: ";
   cin>>e;
   Assign(T,p,e);
   cout<<"先序遍歷二叉樹:"<<endl;
   PreOrderTraverse(T,visit);
   cout<<"結點"<<e<<"的雙親為"<<Parent(T,e)<<",左右孩子分別為";
   cout<<LeftChild(T,e)<<","<<RightChild(T,e)<<",左右兄弟分別為";
   cout<<LeftSibling(T,e)<<","<<RightSibling(T,e)<<endl;
   InitBiTree(s);
   cout<<"建立右子樹為空的樹s:"<<endl;
   CreateBiTree(s);
   cout<<"樹s插到樹T中,請輸入樹T中樹s的雙親結點 s為左(0)或右(1)子樹: ";
   cin>>e>>j;
   InsertChild(T,e,j,s);
   Print(T);
   cout<<"刪除子樹,請輸入待刪除子樹根結點的層號 本層序號 左(0)或右(1)子樹: ";
   cin>>p.level>>p.order>>j;
   DeleteChild(T,p,j);
   Print(T);
   ClearBiTree(T);
   cout<<"清除二叉樹后,樹空否？"<<BiTreeEmpty(T)<<"(1:是 0:否) 樹的深度="<<BiTreeDepth(T)<<endl;
   i=Root(T,e);
   if(i)
     cout<<"二叉樹的根為："<<e<<endl;
   else
     cout<<"樹空，無根"<<endl;
 }