我有一个家庭作业要求我创建一个二叉搜索树的结构，其二叉搜索树的节点是另一个二叉搜索树。第一个BST有学生姓氏，另一个有名字和姓名。此外，如果某人与另一名学生姓氏相同，我不得创建另一个“姓氏”节点，但我必须在现有的“姓氏”节点内创建另一个“名字和身份”节点。更具体一点：
typedef struct nameANDid{ //name and id nodes
char first[20];
int ID;
struct nameANDid *nleft;
struct nameANDid *nright;
}yohoho;
typedef struct node{ //surname nodes
char last[20];
struct nameANDid yohoho;
struct node *left;
struct node *right;
}node;
我的主要问题是如何为我找到的每个名字创建一个不同的nameANDid节点，因为我使用以下代码创建了2个BST，一个用于姓氏，另一个用于名称，但我希望如下： 如果我有这些学生
Stallone Sylvester 11111111
Stallone Noah 22222222
Norris Chuck 33333333
Hogan Hulk 44444444
Hogan Daniel 55555555
我想像这样存储它们：.........
Stallone Sylvester 11111111
Noah 22222222
Norris Chuck 33333333
Hogan Hulk 44444444
Daniel 55555555
取而代之的是我采取的措施：...........
Stallone Sylvester 11111111.
Noah 22222222
Chuck 33333333
Hulk 44444444
Daniel 55555555
Norris Sylvester 11111111.
Noah 22222222
Chuck 33333333
Hulk 44444444
Daniel 55555555
Hogan Sylvester 11111111.
Noah 22222222
Chuck 33333333
Hulk 44444444
Daniel 55555555
为了更具体，我会在这里添加一些功能
load函数从txt文档中加载名称。
void loadData(struct node *temp){
int i;
FILE *fp;
fp=fopen(FILENAME,"r");
if (fp == NULL) printf("File does not exist\n");
for (i=0; i<5; i++){
fscanf(fp,"%s",&temp->last);
fscanf(fp,"%s",&temp->yohoho.first);
fscanf(fp,"%d",&temp->yohoho.ID);
top=add_node(top,temp); //this function create a surname node
}
fclose(fp);
printf("\n\nFile loaded\n");
}
其中
struct node temp;//just a node pointer
struct node *top=NULL; //shows the top of the tree
addnode函数是：...
struct node * add_node (struct node *top, struct node *temp){
struct node *newNode;
if (top == NULL){
newNode=(struct node *)malloc(sizeof(struct node));
temp->left=NULL;
temp->right=NULL;
if (memcpy(newNode,temp,sizeof(struct node)) == NULL){
printf("Node addition failed\n");
return NULL;}
else {
topname=add_node_nameANDid(topname,&temp->yohoho); //Call the add_node_nameANDid to create a new name node in the other tree
return newNode;}
}
else {
if (stricmp(temp->last,top->last) < 0){ //Insert node surname left
top->left=add_node(top->left,temp);}
else if (stricmp(temp->last,top->last) == 0){
topname=add_node_nameANDid(topname,&temp->yohoho); //Call the add_node_nameANDid to create a new name node in the other tree if i have the same surname
}
else {
top->right=add_node(top->right,temp);
}
return top;
}
return NULL;
}
add_node_nameANDid（）函数与上一个函数类似，但它改变了一些变量：
struct nameANDid * add_node_nameANDid (struct nameANDid *topname, struct nameANDid *temp2){
struct nameANDid *newNode_nameANDid;
if (topname == NULL){
newNode_nameANDid=(struct nameANDid *)malloc(sizeof(struct nameANDid));
temp2->nleft=NULL;
temp2->nright=NULL;
if (memcpy(newNode_nameANDid,temp2,sizeof(struct nameANDid)) == NULL){
printf("Node addition failed\n");
return NULL;}
else {
return newNode_nameANDid;}
}
else {
if (stricmp(temp2->first,topname->first) <= 0){
topname->nleft=add_node_nameANDid(topname->nleft,temp2);}
else {
topname->nright=add_node_nameANDid(topname->nright,temp2);}
return topname;
}
return NULL;
}
很抱歉我刚刚上传了庞大的源代码，但如果没有这个代码就很难解释。
我认为我有两个问题，但我没有解决问题的知识。
第一：我必须为每个姓氏节点创建不同的名字BST，我认为我不这样做，但我不知道该怎么做......
有什么建议吗？
答案 0 :(得分：2)
我在下面给出了一个示例实现，评论说明我是如何解决这个问题的。您应该能够使用我的想法来修改代码的工作方式。请注意，它不是一个完美的实现，在我的头脑中，我可以看到以下问题。
for / while循环而不是调用自身的函数 - 这将允许您的机器内存可以处理的节点数量很多（修复问题）。add_name_to_tree以处理balanced binary tree的插入（但这只会解决问题，堆栈限制仍然存在）。我将把它作为练习来处理这些情况。
#include <stdio.h>
#include <string.h>
/* a single struct type for storing all tree elements */
typedef struct _node
{
char name[50];
int id;
struct _node *subname;
struct _node *left;
struct _node *right;
} node;
/* creates a new node structure for the specified name and id */
node *create_node(const char *name, int id)
{
node *newNode = (node*)malloc(sizeof(node));
memset(newNode, 0, sizeof(*newNode));
newNode->id = id;
strncpy(newNode->name, name, sizeof(newNode->name));
return newNode;
}
/* inserts the name/id pair into the tree specified by root.
note that root is passed as a pointer to a pointer, so that
it can accept NULL if no tree exists yet, and return to the
caller the node the node that contains the name. Note that
id is ignored if "name" already exists, i'll leave it as an
excersice for you to handle situations with the same name
with multiple id's */
node *add_name_to_tree(node **root, const char *name, int id)
{
if (*root == NULL)
{
*root = create_node(name, id);
return *root;
}
const int cmp = strcmp(name, (*root)->name);
if (cmp < 0)
{
return add_name_to_tree(&(*root)->left, name, id);
}
else if (cmp > 0)
{
return add_name_to_tree(&(*root)->right, name, id);
}
else
{
return *root;
}
}
/* adds the specified first/last name and id combo to the tree
specified by root */
node *add_name(node *root, const char *first, const char *last, int id)
{
/* this call will return the node that holds the last name,
we can then use its "subname" tree root to insert the first name */
node *last_node = add_name_to_tree(&root, last, 0);
/* use the "subname" of the node that stores the last name as the
root of the tree that stores first names */
add_name_to_tree(&last_node->subname, first, id);
return root;
}
/* just to demonstrate why I use the same node type for first/last names,
its because it allows you to support any number of names, see
below - an add function that adds people with a middle name to the tree
*/
node *add_with_middle_name(node *root, const char *first,
const char *middle, const char *last, int id)
{
node *last_node = add_name_to_tree(&root, last, 0);
node *mid_node = add_name_to_tree(&last_node->subname, middle, 0);
add_name_to_tree(&mid_node->subname, first, id);
return root;
}
/* recursively traverse the name tree, printing out the names */
void print_names(node *names, int level)
{
const int indent = 10;
if (names == NULL)
{
printf("\n");
}
if (names->left)
{
print_names(names->left, level);
}
if (names->subname)
{
printf("%*c %s \n", (indent * level), ' ', names->name);
print_names(names->subname, level + 1);
printf("\n");
}
else
{
printf("%*c %-*s %d\n",
(indent * level), ' ',
indent, names->name, names->id);
}
if (names->right)
{
print_names(names->right, level);
}
}
int main()
{
node *names = NULL;
names = add_name(names, "Sylvester", "Stallone", 11111111);
names = add_name(names, "Noah", "Stallone", 22222222);
names = add_name(names, "Chuck", "Norris", 33333333);
names = add_name(names, "Hulk", "Hogan", 44444444);
names = add_name(names, "Daniel", "Hogan", 55555555);
names = add_with_middle_name(names, "Peter", "Michael",
"Zachson", 66666666);
print_names(names, 0);
return 0;
}