怎样将二叉树的线索化,该怎么处理

怎样将二叉树的线索化
rt

[解决办法]
这个呀，我以前写个几个 ，你自己参考参考
#include "iostream.h"
typedef char TelemType;
typedef struct BiTNode
{
TelemType data;
struct BiTNode *lchild,*rchild; /*左右孩子指针*/
int ltag,rtag; 
 }BiTNode,*BiTree;

BiTree pre=0;

//按先序遍历得到的字符串建立二叉链表
BiTNode *createbtree()
{BiTNode *t;
TelemType ch;
cin>>ch;
if(ch=='#') return 0;
else
{t=new BiTNode;
t->data=ch;
t->ltag=t->rtag=0;
t->lchild=createbtree();
t->rchild=createbtree();
}
return t;
}
//先序遍历
void preorder(BiTNode *t)
{if(t!=0)
{
cout<<t->data<<" ";

preorder(t->lchild);
preorder(t->rchild);
}
}
//线索化
void PreThreading(BiTNode *P)
{ if (P)
{ if (!P->lchild)
{ P->ltag=1;P->lchild=pre;}
if (!P->rchild) P->rtag=1; 
if(pre && pre->rtag==1)
pre->rchild=P;
pre=P;
 if (P->ltag==0)
PreThreading(P->lchild);
if (P->rtag==0) PreThreading(P->rchild);
}
}

//查找
BiTNode *locate(BiTNode *bt,TelemType x)
{BiTNode *p;
if(bt==0) return 0;
if(bt->data==x)return bt;
if(bt->lchild !=0) 
{p=locate(bt->lchild,x);
if(p!=0) return p;
}
if(bt->rchild !=0)
{p=locate(bt->rchild,x);
if(p!=0) return p;
}
return 0;
}
//输出二叉树（向左旋转90度）
void printbtree(BiTNode *t,int level)
{ if(t!=0)
{ printbtree(t->rchild,level+1);
if(level!=0)
{ for(int i=0;i<6*(level-1);i++) cout<<" ";
cout<<"----";
}
cout<<t->data<<endl;
printbtree(t->lchild ,level+1);
}
}
//先序线索二叉树中查找结点t的后继
BiTree preordernext(BiTree t)
{
if(t->lchild) return t->lchild;
if(t->rchild) return t->rchild;
return 0;

/*if(t->rtag==1) return (t->rchild);
else
{ p=t->rchild;
while(p->ltag==0) p=p->lchild;
if(p==t) return null;
return p;
}*/
}
//先序线索二叉树中查找结点p的前驱
BiTree preorderpre(BiTree t,BiTree p)
{
if(t==p) return pre;
pre=p;
 if(t->lchild) preorderpre(t->lchild,p);
 if(t->rchild) preorderpre(t->rchild,p);
 
}


void main()
{ int x=0;
BiTNode *t1,*t2=0,*t3;
t1=createbtree(); 
printbtree(t1,x); 
t2=locate(t1,'d');
preorder(t1);cout<<endl;
PreThreading(t1);
t3=preordernext(t2);
if(t3)cout<<t3->data <<endl;
t3=preorderpre(t1,t2);
if(t3) cout<<t3->data <<endl;
}