线索化二叉树的相关操作

Posted 2020-07-07 Kooing

#include <iostream>

using namespace std;

typedef struct Tree
{
    char data;
    int lTag, rTag;
    struct Tree *lchild, *rchild;
}Tree;

void firstCreateTree(Tree* &T)
{
    char ch;
    cin >> ch;
    if (‘#‘ == ch)
        T= NULL;
    else
    {
        T = new Tree;
        T->data=ch;
        firstCreateTree(T->lchild);
        firstCreateTree(T->rchild);
    }
}

Tree* pre;

void InThreading(Tree* &T)
{//树线索化
    if (T)
    {
        InThreading(T->lchild);
        if (!T->lchild)
        {
            T->lTag = 1;
            T->lchild = pre;
        }
        else
            T->lTag = 0;
        if (!pre->rchild)
        {
            pre->rTag = 1;//这个很关键，补回去前面的rTag
            pre->rchild = T;
        }
        else
            T->rTag = 0;//不是pre，而是T
        pre = T;
        InThreading(T->rchild);
    }
}

void InOrderThreading(Tree* &T, Tree* &Thrt)
{//整棵树线索化
    Thrt = new Tree;
//    Thrt->lchild = T;
    Thrt->rTag = 1;
    Thrt->rchild = Thrt;
    Thrt->lTag = 0;
    if (!T)Thrt->lchild = Thrt;
    else
    {
        Thrt->lchild = T;
        pre = Thrt;
        InThreading(T);
        pre->rchild = Thrt;
        pre->rTag = 1;
        Thrt->rchild = pre;
    }
}

void InorderTraverse(Tree* T)
{
    Tree *p = T->lchild;
    for (; p != T;)
    {
        for (; p->lTag == 0;)p = p->lchild;//往下搞到1，也就是没有左孩子
        cout << p->data;                  //输出最左的一个孩子
        for (; p->rTag == 1 && p->rchild != T;)//看右孩子是不是为1，也就是空
        {
            p = p->rchild;            //空的话跟线索回到上一个并输出
            cout << p->data;
        }
        p = p->rchild;             //不空的话就到了右孩子那里，看起来和上面的相同，其实是不一样的
    }
}

int main()
{
    Tree *T,*Thrt;
    firstCreateTree(T);
    InOrderThreading(T,Thrt);
//    cout << Thrt->rTag << endl;曾经不会这个点
    InorderTraverse(Thrt);
}