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headvoid 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.outABC__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 0x1a83770self 0x1a83770 - lc (nil), lt 0, D, rt 0, rc (nil)self 0x1a836e0 - lc 0x1a83710, lt 0, A, rt 0, rc 0x1a837a0self 0x1a837a0 - lc (nil), lt 0, E, rt 0, rc 0x1a837d0self 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 0x1a83770pre 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 0x1a837a0pre p 0x1a836e0 - lc 0x1a83710, lt 0, A, rt 0, rc 0x1a837a0 inThread p 0x1a837a0 - lc 0x1a836e0, lt 1, E, rt 0, rc 0x1a837d0pre 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 0x1a837d0self 0x1a83740 - lc 0x1a832a0, lt 1, C, rt 1, rc 0x1a83710self 0x1a83710 - lc 0x1a83740, lt 0, B, rt 0, rc 0x1a83770self 0x1a83770 - lc 0x1a83710, lt 1, D, rt 1, rc 0x1a836e0self 0x1a836e0 - lc 0x1a83710, lt 0, A, rt 0, rc 0x1a837a0self 0x1a837a0 - lc 0x1a836e0, lt 1, E, rt 0, rc 0x1a837d0self 0x1a837d0 - lc 0x1a837a0, lt 1, F, rt 0, rc (nil)[root@8be225462e66 c]#