关于线索 二叉树，为什么只遍历了根节点的左子树？？？
二叉树 遍历 c C++ 算法
#include<iostream>
#include<stdlib.h>
using namespace std;
typedef char ElemType;
typedef enum{link,thread}PointerTag;//link==0：指针，thread==1：线索
typedef struct TNode
{
ElemType data;//数据域
struct TNode *lch,*rch;//左右孩子指针
PointerTag LTag,RTag;//标志域
}TNode,*Bitree;
Bitree pre;
Bitree CreatT(Bitree &T)//创建二叉树
{
ElemType ch;
cout<<" ";
cin>>ch;
if(ch=='#')
T = NULL;
else
{
if(!(T=new TNode))
exit(OVERFLOW);
T->data = ch;
CreatT(T->lch);
CreatT(T->rch);
}
return T;
}
void InThreading(Bitree p)//中序遍历进行中序线索化
{
if(p)
{
InThreading(p->lch);
if(!p->lch)
{
p->LTag = thread;
p->lch = pre;
}
else
p->LTag = link;
if(!pre->rch)
{
pre->RTag = thread;
pre->rch = p;
}
else
p->RTag = link;
pre = p;
InThreading(p->rch);
}
}
void InOrderThreading(Bitree &thr,Bitree T)//建立头结点，中序线索二叉树
{
if(!(thr=new TNode))
exit(OVERFLOW);
thr->LTag = link;
thr->RTag = thread;
thr->rch = thr;
//thr->RTag = link;
if(!T)
thr->lch = thr;
else
{
thr->lch = T;
pre = thr;
InThreading(T);
pre->rch = thr;
pre->RTag = thread;
thr->rch = pre;
}
}
void InOrderTraverse(Bitree T)
{
Bitree p = T->lch;
while(p!=T)//空树或遍历结束时，p==T
{
while(p->LTag==link)
p = p->lch;
cout<<p->data<<" ";
while(p->RTag==thread&&p->rch!=T)
{
p = p->rch;
cout<<p->data<<" ";
}
p = p->rch;
}
}
int main()
{
Bitree T,thr;
cout<<"Enter data of binary tree:\n";//输入ABDH###E##CF##G##
CreatT(T);
InOrderThreading(thr,T);
cout<<"Print data of binary tree:\n";//只输出了根节点的左子树HDBE
InOrderTraverse(T);
cout<<endl;
return 0;
}
