【问题描述】

以树形结构实现域名的搜索，即输入某站点的域名，在域名系统的树形结构中进行搜索，直至域名全部全部匹配成功或者匹配失败，若成功则给出该节点的IP地址，否则给出找不到该节点信息。

【基本要求】

首先实现一个反映域名结构的树，如清华大学站点为www.，在该树从根到叶子的各层节点就应该是root、cn、edu、tsinghua、www。叶子结点www另有一个数据域，存放清华大学站点的IP地址166.111.9.2

#include <stdio.h>#include <stdlib.h>#include <string.h>typedef struct app {char a[20];struct app *left;struct app *right;char b[20];} app;int cout = 0;/*建树时用到的数据：com baidu www 220.181.27.5 # # google www 66.249.89.104 # #microsoft www 207.46.20.60 # # # gov whitehouse www 64.215.166.127 # #nasa www 210.254.57.56 # # # cn com lenovo www 219.239.195.11 # #sina www 218.30.13.51 # # # edu ustc www 202.38.64.2 # bbs 202.38.64.3 # #pku www 162.105.129.12 # bbs 162.105.204.150 # # tsinghua www 166.111.4.100 #ftp 166.111.8.229 # # # gov beijing www 210.73.64.10 # #shanghai www 61.129.65.58 # # # # #*///数据保存在相同目录下的text.txt文件下;app *jianshu(app *bt, FILE *fp) {char a[20];if (fscanf(fp, "%s ", &a) != EOF) {if (a[0] == '#')bt = NULL;else {bt = (app *)malloc(sizeof(app));strcpy(bt->a, a);if (strcmp(a, "www") == 0 || strcmp(a, "bbs") == 0 || strcmp(a, "ftp") == 0 ) {fscanf(fp, "%s ", &a);strcpy(bt->b, a);}bt->left = jianshu(bt->left, fp);bt->right = jianshu(bt->right, fp);}}return bt;}app *tianjia(app *bt, char fenge[4][20], int i, char ip[20]) {/*puts(fenge[i]);puts(bt->a);printf("%d", i);*/if (bt != NULL && strcmp(bt->a, fenge[i]) == 0) {if (i == 0) {puts(bt->b);return bt;}bt->left = tianjia(bt->left, fenge, --i, ip);} else {if (bt == NULL) {if (i == -1) {return bt;}bt = (app *)malloc(sizeof(app));bt->right = NULL;bt->left = NULL;strcpy(bt->a, fenge[i]);if (strcmp(fenge[i], "www") == 0 || strcmp(fenge[i], "bbs") == 0 || strcmp(fenge[i], "ftp") == 0 ) {strcpy(bt->b, ip);}bt->left = tianjia(bt->left, fenge, --i, ip);} elsebt->right = tianjia(bt->right, fenge, i, ip);}return bt;}void qianxu(app *bt, FILE *fp) {if (bt == NULL) {fprintf(fp, "# ");} else {fprintf(fp, "%s ", bt->a);if (strcmp(bt->a, "www") == 0 || strcmp(bt->a, "bbs") == 0 || strcmp(bt->a, "ftp") == 0 ) {fprintf(fp, "%s ", bt->b);}qianxu(bt->left, fp);qianxu(bt->right, fp);}}void jiediangeshu(app *bt, char fenge[4][20], int i) {/*for (i; i >= 0; i--) {puts(fenge[i]);}*///puts(bt->a);//puts(fenge[i]);if (bt != NULL && strcmp(bt->a, fenge[i]) == 0) {if (i == 0) {puts(bt->b);return;}jiediangeshu(bt->left, fenge, --i);} else {if (bt == NULL) {printf("找不到服务器或发生 DNS 错误");return;}jiediangeshu(bt->right, fenge, i);}}int main() {printf("欢迎使用IP查询!\n");char str[80];char ip[20];char *token;char ju = 'y';int xuan;char fenge[4][20];int flag = 0;app *bt;FILE *fp1 = NULL;FILE *fp2 = NULL;fp1 = fopen("text.txt", "r");bt = jianshu(bt, fp1);fp2 = fopen("text.txt", "w");//qianxu(bt);while (ju != 'n') {printf("请选择功能：1，增加网址；2，查询网址;3,退出系统\n");scanf("%d", &xuan);getchar();if (xuan == 3) {qianxu(bt, fp2);exit(0);}printf("请输入地址：");scanf("%s", &str);getchar();int i = 0;token = strtok(str, ".");while (token != NULL) {strcpy(fenge[i], token);i++;token = strtok(NULL, ".");}switch (xuan) {case 1:printf("请输入ip：");scanf("%s", &ip);getchar();bt = tianjia(bt, fenge, --i, ip);break;case 2:jiediangeshu(bt, fenge, --i);break;}printf("是否继续？y/n\n");scanf("%c", &ju);//qianxu(bt, fp2);getchar();}qianxu(bt, fp2);fclose(fp2);fclose(fp1);printf("谢谢使用");}/*测试数据 220.181.27. 66.249.89. 207.46.20.60www.whitehouse.gov 64.215.166.127www.nasa.gov 210.254.57. 219.239.195. 218.30.13.51www. 202.38.64.2bbs. 202.38.64.3www. 162.105.129.12bbs. 162.105.204.150www. 166.111.4.100ftp. 166.111.8.229www. 210.73.64.10www. 61.129.65.58*/

二叉树结构如图