线索二叉树（C语言）

实现下面这棵树：
先序遍历: A B C D E F
中序遍历: C B D A E F

代码

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <unistd.h>

typedef enum {links, thread} TAG;

typedef struct treeNode {
    char name;
    struct treeNode *lchild, *rchild;
    TAG ltag;
    TAG rtag;
}TREENODE, *TREE;

void createTree(TREE *);
void traverse(TREE);
void traverse_middle(TREE);
void traverse_middle_detail(TREE);
// 线索化二叉树,相比普通的中序遍历，这里把输出节点数据的步骤改成了判断指针域的逻辑
void inThread(TREE, TREE *, TREE);
void traverse_inThread_by_rchild(TREE);

// 调用此函数时须要传入head
void traverse_inThread_by_rchild(TREE head)
{
    printf("中序正向遍历二叉链表:\n");
    printf("self %p - lc %10p, lt %u, %c, rt %u, rc %10p\n", head, head->lchild, head->ltag, head->name, head->rtag, head->rchild);
    TREE p = head->lchild;
    while (p->lchild != head) {
        p = p->lchild;
        //usleep(100000);
    }

    while (p) {
        printf("self %p - lc %10p, lt %u, %c, rt %u, rc %10p\n", p, p->lchild, p->ltag, p->name, p->rtag, p->rchild);
        p = p->rchild;
    }

}

void inThread(TREE p, TREE *pre, TREE head)
{
    if (! p)
        return;

    inThread(p->lchild, pre, head);
        printf("pre p %p - lc %10p, lt %u, %c, rt %u, rc %10p\n", (*pre), (*pre)->lchild, (*pre)->ltag, (*pre)->name, (*pre)->rtag, (*pre)->rchild);

    // 判断自身是否有左孩子，若是没有指向前驱节点
    if (! p->lchild) {
        p->ltag = thread;
        p->lchild = *pre;
    }

    /*
     * 由于遍历(中序)时，路径只走到当前节点，并不知道后继是否有，
     * 因此每一个节点都只处理本身的前驱和前驱的后继
     *  head节点rchild在第1个节点处理时指向了第1个节点
     */
    if (! (*pre)->rchild) {
        (*pre)->rtag = thread;
        (*pre)->rchild = p;
    }
        printf("    inThread p %p - lc %10p, lt %u, %c, rt %u, rc %10p\n", p, p->lchild, p->ltag, p->name, p->rtag, p->rchild);

    // 本节点处理完后，更新pre指向自身，做为中序遍历下一个节点的前驱
    *pre = p;
    // 头指针rchild指向当前节点，最终线索化完成后，头节点的右孩子一定指向中序最后1个节点
    head->rchild = p;

    inThread(p->rchild, pre, head);
}

void traverse_middle(TREE p)
{
    if (p) {
        traverse_middle(p->lchild);
        printf("%c ", p->name);
        traverse_middle(p->rchild);
    }
}

void traverse_middle_detail(TREE p)
{
    if (p) {
        traverse_middle_detail(p->lchild);
        printf("self %p - lc %10p, lt %u, %c, rt %u, rc %10p\n", p, p->lchild, p->ltag, p->name, p->rtag, p->rchild);
        traverse_middle_detail(p->rchild);
    }
}

void traverse(TREE p)
{
    if (p) {
        printf("%c ", p->name);
        traverse(p->lchild);
        traverse(p->rchild);
    }
}

// 前序初始化树的各节点
void createTree(TREE *p)
{
    char c;

    scanf("%c", &c);

    if (c == '_') {
        *p = NULL;
    }
    else {
        *p = (TREE)malloc(sizeof(TREENODE));
        (*p)->name = c;

        // 不管是否会有左右孩子，都先把tag标识为links
        (*p)->ltag = (*p)->rtag = links;

        createTree(&(*p)->lchild);
        createTree(&(*p)->rchild);
    }

}

int main(void)
{
    // 头指针，指向线索二叉树的头节点(该节点的lchild指向root)
    TREE head = NULL;
    TREE tree;

    head = (TREE)malloc(sizeof(TREENODE));
    head->lchild = head->rchild = NULL;
    head->ltag = head->rtag = thread;
    // 为了方便确认头节点
    head->name = 'H';

    TREE pre = head;

    createTree(&tree);
    // 头节点lchild手动指向tree根节点(rchild已经在线索化完成后指向了中序最后1个节点)
    head->lchild = tree;

    printf("先序遍历: ");
    traverse(tree);
    putchar('\n');
    printf("中序遍历: ");
    traverse_middle(tree);
    putchar('\n');
    printf("中序遍历(detail):\n");
    traverse_middle_detail(tree);
    putchar('\n');

    // 线索化二叉树(把空闲的lchild, rchild指向各自的前驱和后继) 
    inThread(tree, &pre, head);

    // 使用rchild遍历中序线索化的二叉链表
    traverse_inThread_by_rchild(head);

    /*
     * 目前中序最后1个节点的rchild依然是NULL，可是已经能够实现根据头节点正反向遍历二叉链表
     * 若是按照其它教程里的须要把中序尾节点rchild的指向头节点，则中序遍历记住最后1个指针操做一下就能够。。。(若是须要判断空树等状况能够参考网上其它教程)
     */

    return 0;
}

output

[root@8be225462e66 c]# gcc thrTree.c && ./a.out
ABC__D__E_F__
先序遍历: A B C D E F
中序遍历: C B D A E F
中序遍历(detail):
self 0x1a83740 - lc      (nil), lt 0, C, rt 0, rc      (nil)
self 0x1a83710 - lc  0x1a83740, lt 0, B, rt 0, rc  0x1a83770
self 0x1a83770 - lc      (nil), lt 0, D, rt 0, rc      (nil)
self 0x1a836e0 - lc  0x1a83710, lt 0, A, rt 0, rc  0x1a837a0
self 0x1a837a0 - lc      (nil), lt 0, E, rt 0, rc  0x1a837d0
self 0x1a837d0 - lc      (nil), lt 0, F, rt 0, rc      (nil)

pre p 0x1a832a0 - lc  0x1a836e0, lt 1, H, rt 1, rc      (nil)
    inThread p 0x1a83740 - lc  0x1a832a0, lt 1, C, rt 0, rc      (nil)
pre p 0x1a83740 - lc  0x1a832a0, lt 1, C, rt 0, rc      (nil)
    inThread p 0x1a83710 - lc  0x1a83740, lt 0, B, rt 0, rc  0x1a83770
pre p 0x1a83710 - lc  0x1a83740, lt 0, B, rt 0, rc  0x1a83770
    inThread p 0x1a83770 - lc  0x1a83710, lt 1, D, rt 0, rc      (nil)
pre p 0x1a83770 - lc  0x1a83710, lt 1, D, rt 0, rc      (nil)
    inThread p 0x1a836e0 - lc  0x1a83710, lt 0, A, rt 0, rc  0x1a837a0
pre p 0x1a836e0 - lc  0x1a83710, lt 0, A, rt 0, rc  0x1a837a0
    inThread p 0x1a837a0 - lc  0x1a836e0, lt 1, E, rt 0, rc  0x1a837d0
pre p 0x1a837a0 - lc  0x1a836e0, lt 1, E, rt 0, rc  0x1a837d0
    inThread p 0x1a837d0 - lc  0x1a837a0, lt 1, F, rt 0, rc      (nil)
中序正向遍历二叉链表:
self 0x1a832a0 - lc  0x1a836e0, lt 1, H, rt 1, rc  0x1a837d0
self 0x1a83740 - lc  0x1a832a0, lt 1, C, rt 1, rc  0x1a83710
self 0x1a83710 - lc  0x1a83740, lt 0, B, rt 0, rc  0x1a83770
self 0x1a83770 - lc  0x1a83710, lt 1, D, rt 1, rc  0x1a836e0
self 0x1a836e0 - lc  0x1a83710, lt 0, A, rt 0, rc  0x1a837a0
self 0x1a837a0 - lc  0x1a836e0, lt 1, E, rt 0, rc  0x1a837d0
self 0x1a837d0 - lc  0x1a837a0, lt 1, F, rt 0, rc      (nil)
[root@8be225462e66 c]#