先序 中序 后序 线索二叉树

///有头结点版 + 无头结点版  啊啊啊啊啊啊 感觉自己快吐血了  T^T#include<iostream>#include<queue>using namespace std;typedef struct treenode{    char data;    treenode *lchild;    treenode *rchild;    int LTag;    int RTag;}node, *Tree;node *pre;node *create_tree(node *t){    char a;    cin >> a;    if(a == '0')        return NULL;    t = new node;    t->LTag = 0;    t->RTag = 0;    t->data = a;    t->lchild = create_tree(t->lchild);    t->rchild = create_tree(t->rchild);    return t;}void pre_order(node *t){    if(t){        cout << t->data;        pre_order(t->lchild);        pre_order(t->rchild);    }}///******************in_order**************************void in_order_threading(Tree t){    if(t){        in_order_threading(t->lchild);        if(t->lchild == NULL){            t->lchild = pre;            t->LTag = 1;        }        if(pre && pre->rchild == NULL){            pre->rchild = t;            pre->RTag = 1;        }        pre = t;        in_order_threading(t->rchild);    }}node * in_thread(node *p, Tree t){    p = new node;    p->LTag = 0;    p->RTag = 1;    p->rchild = p;    if(t == NULL){        p->lchild = p;    }    else{        p->lchild = t;        pre = p;        in_order_threading(t);        p->rchild = pre;        pre->RTag = 1;        pre->rchild = p;    }    return p;}void in_order_traverse(node *p){    node *t;    t = p->lchild;    while(t!=p){        while(t->LTag == 0){///走到最左             t = t->lchild;        }        cout << t->data;///打印最左边的        while(t->RTag == 1 && t->rchild!=p){            t = t->rchild;///顺着线索去打印             cout << t->data;        }        t = t->rchild;    }}///**************pre_order*************************void pre_order_threading(Tree t){    if(t){        if(t->lchild==NULL){            t->lchild = pre;            t->LTag = 1;        }        if(pre && pre->rchild == NULL){            pre->rchild = t;            pre->RTag = 1;        }        pre = t;        if(t->LTag == 0)/// 之前写的是t->lchild 死循环了  这两个不是等价吗？？？        pre_order_threading(t->lchild);///终于在写后序的时候反应过来了。。。        if(t->RTag == 0)///因为我们已经把空的t->lchild赋值了 其他的是没关系的        pre_order_threading(t->rchild);///但是有一个是NULL 就是说 t->LTag == 1    }///但是t->lchild == NULL 也就是说 只有t->LTag==0 t->lchild才一定不空    return;///不管你明不明白 反正我是明白了~ 唉 困扰了我一下午的问题啊 竟然在一瞬间就有了灵感~}node *pre_thread(node *p, Tree t){    p = new node;    p->LTag = 0;    p->RTag = 1;    p->rchild = p;    if(t==NULL){        p->lchild = p;    }    else{        p->lchild = t;        pre = p;        pre_order_threading(t);        p->rchild = pre;        pre->RTag = 1;///现在pre是最后一个节点啦        pre->rchild = p;    }    return p;}void pre_order_traverse(node *p){    node *t;    t = p->lchild;    while(t!=p){        while(t->LTag == 0){            cout << t->data;            t = t->lchild;        }        cout << t->data;        while(t->RTag == 1 && t->rchild!=p){            t = t->rchild;            cout << t->data;        }        t = t->rchild;    }}///**************post_order*********************void post_order_threading(Tree t){    if(t){            post_order_threading(t->lchild);            post_order_threading(t->rchild);        if(t->lchild == NULL){            t->lchild = pre;            t->LTag = 1;        }        if(pre&&pre->rchild == NULL){            pre->rchild = t;            pre->RTag = 1;        }        pre = t;    }}node *parent(node *t, node *p){    if(t){        queue<node *>q;        while(!q.empty())            q.pop();        q.push(t);        while(!q.empty()){            node *a = q.front();            q.pop();            if(a->lchild == p || a->rchild == p){                return a;            }            if(a->lchild)                q.push(a->lchild);            if(a->rchild)                q.push(a->rchild);        }    }    return NULL;}void  post_order_traverse(node* t){    if (t)    {        node* p = t;        pre = NULL;        while (p)        {            while ( p->lchild != pre && p->LTag == 0)///走到最左            {                p = p->lchild;            }            /// p->data == d            while (p && p->RTag== 1)///顺着线索走            {                cout << p->data;                pre = p;                p = p->rchild;            }            if (p == t)///若p回到了根节点            {                cout << p->data;                return;            }            while (p && p->rchild == pre)            {                cout << p->data;                pre = p;                node *s = parent(t, p);                p = s;            }            if (p && p->RTag == 0)            {                p = p->rchild;            }        }        }}int main(){    node *t;    node *p;    t = create_tree(t);    cout << "先序遍历：";    pre_order(t);    cout << endl;    cout << "中序线索化：";    p = in_thread(p, t);    in_order_traverse(p);    cout << endl;     t = create_tree(t);    cout << "先序线索化：";    p = pre_thread(p, t);    pre_order_traverse(p);    cout << endl;    t = create_tree(t);    cout << "后序线索化：";    post_order_threading(t);    post_order_traverse(t);    cout << endl;    return 0;}/*ab0d00ce000先序遍历：abdce中序线索化：bdaecab0d00ce000先序线索化：abdceab0d00ce000后序线索化：dbecaProcess returned 0 (0x0)   execution time : 22.250 sPress any key to continue.*/