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

#include

#include

#include

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为了判断指针域的逻辑

void inThread(TREE, TREE *, TREE);

void traverse_inThread_by_rchild(TREE);

// 调用此函数时需要传入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，如果没有指向前驱节点

if (! p->lchild) {

p->ltag = thread;

p->lchild = *pre;

}

/*

* 因为遍历(中序)时，路径只走到当前节点，并不知道后继是否有，

* 所以每个节点都只处理自己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，作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，头节点的右孩子必定指向中序最后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;

// 无论是否会有左右孩子，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，但是已经可以实现根据头节点正反向遍历二叉链表

* 如果按照其它教程里的需要把中序尾节点rchild的指向头节点，则中序遍历记住最后1个指针操作一下就可以。。。(如果需要判断空树等情况可以参考网上其它教程)

*/

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]#