I've been stuck on the insertion part of the binary search tree. I get so confused with nested structs. The basic idea of this program is to create a bst that is able to hold names and double values which get stored by value (obviously).
Example: I want to store
Jane 3.14
John 3.233
Luke 6.4
Mike 1.4
so the bst would look like
3.14
/ \
1.4 3.233
\
6.4
however I'm having trouble with the insertHelper recursion portion of the code. The hash table is a bonus part of the code that I'll try implementing at a later time. Thank you for your help!
typedef struct name_val // holds name and value
{
char *name;
double value;
}NAME_VAL;
typedef struct node //binary search tree
{
NAME_VAL *nV;
struct node *left;
struct node *right;
}NODE;
struct tmap_struct //handle for bst and hashtable
{
int nL; //nodes on left
int nR; //nodes on right
NODE *root;
NODE **table;
};
int tmap_insert(TMAP_PTR hashTree, char * name, double val)
{
if(hashTree->root == NULL)
{
NODE *bst = (NODE *)malloc(sizeof(NODE));
NAME_VAL *root = (NAME_VAL *)malloc(sizeof(NAME_VAL));
bst->nV = root;
bst->nV->value = val;
strcpy(bst->nV->name, name);
hashTree->root = bst;
hashTree->nL = 0;
hashTree->nR = 0;
}
else
insertHelper(hashTree->root, val, name);
}
void insertHelper(TMAP_PTR hashTree, int val, char * name)
{
if(val < hashTree->root->nV->value)
{
if(hashTree->root->left == NULL)
{
hashTree->root->left = (NODE *)malloc(sizeof(NODE));
hashTree->root->left->nV = (NAME_VAL *) malloc(sizeof(NAME_VAL));
strcpy(hashTree->root->left->nV->name, name);
hashTree->root->nV->value = val;
(hashTree->nL)++;
}
else
insertHelper(hashTree->root->left, val, name);
}
else
{
if(hashTree->root->right == NULL)
{
hashTree->root->right = (NODE *)malloc(sizeof(NODE));
hashTree->root->right->nV = (NAME_VAL *)malloc(sizeof(NAME_VAL));
strcpy(hashTree->root->left->nV->name,name);
hashTree->root->nV->value = val;
(hashTree->nR)++;
}
else
insertHelper(hashTree->root->right, val, name);
}
}
I doubt this compiles. Is that the problem you're having?
From what I can see, you have declared insertHelper with the wrong type for its first parameter. It should take NODE* values, not TMAP_PTR values. That's because you always call it with nodes out of your tree.
So the first part of the function should look like this:
void insertHelper(NODE *node, int val, char * name)
{
if(val < node->nV->value)
{
if(node->left == NULL)
{
node->left = (NODE *)malloc(sizeof(NODE));
node->left->nV = (NAME_VAL *) malloc(sizeof(NAME_VAL));
strcpy(node->left->nV->name, name);
node->left->nV->value = val;
}
else
insertHelper(node->left, val, name);
}
//.....
Note that I removed the line:
(hashTree->nR)++;
It hardly even makes sense to track this information, unless maybe you do it at the node level.
But if you must, you could have insertHelper recursively return a positive or negative value to indicate what side it inserted on. But that doesn't makes sense. What is it on the right of? You may have inserted it on the right of a node that was in the left half of the tree.
If you store this information on each node, you can recursively update the node above as you return from insertHelper. Maybe that's what you were trying to do. Balanced tree implementations do something similar - AVL trees store the maximum depth of the tree at a node and use that to do branch rotations for rebalancing.
You'll have to adapt mine(It's almost standard C besides the unneeded template and class), but it's a similar algorithm: (I believe, I didn't look at any source for my own purposes.)
template<typename T>
class BST {
protected:
typedef struct node_t {
struct node_t * dir[2];
T data;
} node;
node * root;
void insert_node(node * active_node, T data){ //call with node *root;
int next = data < active_node->data ? 0 : 1;
if(active_node->dir[next] == NULL){
active_node->dir[next] = new node;
active_node->dir[next]->dir[0] = NULL;
active_node->dir[next]->dir[1] = NULL;
active_node->data = data;
} else
insert_node(active_node->dir[next], data);
}
public:
BST() : root(new node){root->dir[0] = NULL; root->dir[1] = NULL; root->data = 0;}
~BST(){}
}
Related
I have a doubly linked list and I want to change the direction of the list. For instance:
1 -> 2 -> 3
becomes
3 -> 2 -> 1
I have used a boolean expression in my structure to determine the direction.
typedef struct Node Node;
struct Node {
char character;
Node *link[2];
bool dir;
} direction;
I determine the direction with this format:
Node *p;
p->link[p->dir];
or
Node *p;
p->link[!p->dir];
The issue that I'm having is I want to flip these boolean directions with a method that uses a runtime of O(1). I have attempted to build a function that handles it like this:
//global variable
int d = 0;
//function
void switchStack ( ) {
if (d == 0) {
d = 1;
direction->link[direction->dir] = direction->link[!direction->dir];
else if (d == 1) {
d = 0;
direction->link[direction->dir] = direction->link[!direction->dir];
}
This function doesn't seem to do anything and any other variation I try crashes the program when I call it. Does anybody have any idea on how to use the direction switch properly to reverse the stack with a runtime of 0(1)?
This is prefaced by my/the top comments.
I'm still not sure exactly what you want. But, this is as close as I can divine/guess.
The listadd always appends to the tail of the list (i.e. it ignores dir).
But, listprint honors the dir. It is a model for other functions you may write.
But, I think only listprint needs to look at dir. All other functions can treat a given list as a forward list (i.e. see how listadd works).
#include <stdio.h>
#include <stdlib.h>
typedef struct node Node;
struct node {
Node *links[2];
int value;
};
#define PREV 0
#define NEXT 1
typedef struct {
Node *begend[2];
int dir;
} List;
#define HEAD 0
#define TAIL 1
static inline Node *
listfirst(List *list)
{
return list->begend[list->dir ? TAIL : HEAD];
}
static inline Node *
listlast(List *list)
{
return list->begend[list->dir ? HEAD : TAIL];
}
static inline Node *
nodenext(List *list,Node *cur)
{
return cur->links[list->dir ? PREV : NEXT];
}
static inline Node *
nodeprev(List *list,Node *cur)
{
return cur->links[list->dir ? NEXT : PREV];
}
List *
listnew(void)
{
List *list = calloc(1,sizeof(List));
return list;
}
Node *
nodenew(int value)
{
Node *cur = calloc(1,sizeof(Node));
cur->value = value;
return cur;
}
void
listadd(List *list,int value)
{
Node *node = nodenew(value);
Node *prev = list->begend[TAIL];
do {
if (prev != NULL) {
prev->links[NEXT] = node;
node->links[PREV] = prev;
break;
}
list->begend[HEAD] = node;
} while (0);
list->begend[TAIL] = node;
}
void
listprint(List *list)
{
Node *cur;
for (cur = listfirst(list); cur != NULL; cur = nodenext(list,cur))
printf("%d\n",cur->value);
}
int
main(void)
{
List *list = listnew();
listadd(list,1);
listadd(list,2);
listadd(list,3);
printf("\n");
printf("list in forward direction\n");
listprint(list);
// reverse the list
list->dir = ! list->dir;
printf("\n");
printf("list in reverse direction\n");
listprint(list);
return 0;
}
This is the output of the program:
list in forward direction
1
2
3
list in reverse direction
3
2
1
I am relatively new to C, and have been learning about linked lists with pointers.
I learned that
(*foo).bar is the same ad foo->bar.
foo->bar is used because it is more readable.
Therefore I do not understand why these code snippets behave differently:
1)
void appendCourse(CourseNode** pLL, Course c){
CourseNode * root = *pLL;
CourseNode* last = makeCourseNode(c);
if(root != NULL){
CourseNode node = *root;
while(node.pNext != NULL){
node = *node.pNext;
}
node.pNext = last;
} else {
*pLL = last;
}
}
and
2)
void appendCourse(CourseNode** pLL, Course c){
CourseNode * root = *pLL;
CourseNode* last = makeCourseNode(c);
if(root != NULL){
CourseNode *node = root;
while(node->pNext != NULL){
node = node->pNext;
}
node->pNext = last;
} else {
*pLL = last;
}
}
to me it looks like 1) should behave as if dereferencing first, then member access. Sort of like (*foo).bar
but 1) doesn't seem to work right at all, it can only successfully add the first element.
2) does however add all elements into the linked list.
In case this helps: my structs and other method:
typedef struct CourseNode {
struct CourseNode* pNext;
Course course;
} CourseNode;
typedef struct
{
StudentNode *pWaitlistHead; // Waitlist for this course
char szCourseId[12]; // Course Identifier
char szRoom[15]; // Room number of the course
char szDays[15]; // What days the course will meet, ex: MWF, TR, etc
char szTimes[15]; // Meeting Time, ex: 10:00-11:15am
int iAvailSeats; // Number of available seats in the course
double dFee; // Additional fees for the course
} Course;
CourseNode* makeCourseNode(Course c){
CourseNode * node = malloc(sizeof(CourseNode));
node->pNext = NULL;
node->course = c;
return node;
}
CourseNode node = *root;
while(node.pNext != NULL){
node = *node.pNext;
}
This creates a new CourseNode called node. The value of that new CourseNode is modified, but that has no affect on the linked list.
CourseNode *node = root;
while(node->pNext != NULL){
node = node->pNext;
}
Here, node points to a CourseNode that is on the linked list.
The simplest way to understand the difference is that the first code excerpt creates new CourseNodes. It's like the difference between these two:
int foo (int *i)
{
int *j = i; // j is a pointer to the same int i points to
*j = 2; // this changes the value of the int i points to
int j = *i; // this creates a new int
j = 2; // this changes the value of that new int
}
I'm currently doing an assignment for uni and I need to find the sum of a graph.
To do this I believe I need a linked list that I can use to remember which nodes have been visited. I have the linkedlist working correctly but I can't get a contains function to work. This is the code I have:
struct listnode
{
struct N *val;
struct listnode *next;
};
int contains(struct listnode *head,struct N* value)
{
struct listnode *current = head;
while (current)
{
if ((current -> val) == value)
{
return 1;
}
current = current -> next;
}
return 0;
}
note: N is a node of the graph.
Can anyone see any problems with what I'm doing?
EDIT: contains function should return 1 when N *value is in the list, 0 otherwise
EDIT2:
I have a push function:
void push(struct listnode *head,struct N *value)
{
if (head)
{
struct listnode *current = head;
while (current->next)
{
current = current -> next;
}
current->next = malloc(sizeof(struct listnode*));
current->next->val = value;
current->next->next = NULL;
}
else
{
head = malloc(sizeof(struct listnode*));
if (head)
{
head -> val = value;
head -> next = NULL;
}
else
{
printf("error");
exit(0);
}
}
}
and I want the following line to return 1:
contains(push(visited,p),p);
where p is a pointer to a struct N and visited is my global linked list
EDIT3:
this is my final sum function that I believe should work, but doesnt because of contains.
long sum(struct N *p)
{
if (p)
{
if (contains(visited,p) == 0) //if p hasnt been visited
{
push(visited,p); //make it visited
return (p -> data) + sum(p -> x) + sum(p -> y) + sum(p -> z);
}
else
{
return 0;
}
}
else
{
return 0;
}
}
Your contains function appears to be fine. The issue is that you are always passing a NULL list to it, which is caused by a faulty push function. You need a return in push, or to pass in a pointer with one more level of indirection, so you can assign to head outside of push. One more possible improvement is to notice that no matter what you pass in, the malloc and initialization of a new node is actually the same.
Finally, the main issue, that is really the most likely to cause a segfault is the fact that you are allocating enough space for a pointer to a node, not for the node itself.
Here is an example:
#ifdef BY_INDIRECTION
#define RET_TYPE void
#define IN_TYPE struct listnode **
#else
#define RET_TYPE struct listnode *
#define IN_TYPE struct listnode *
#endif
RET_TYPE push(IN_TYPE head, struct N *value)
{
struct listnode *current, **next;
if(head)
{
for(current = head; current->next; current = current->next) ;
next = &(current->next);
}
else
{
#ifdef BY_INDIRECTION
next = head;
#else
next = &head;
#endif
}
*next = malloc(sizeof(struct listnode));
if(!*next) {
printf("error");
exit(0);
}
(*next)->val = value;
(*next)->next = NULL;
#ifndef BY_INDIRECTION
return head
#endif
}
I have included both suggestions here. If you want to read the one where we use indirection (pass in a listnode ** and have void return), choose the path where BY_INDIRECTION is defined. If you want to have head returned (and pass in just a regular listnode *) read the path where BY_INDIRECTION is not defined.
The latter approach has a return value, so it can be used to write a shortened form like if(contains(push(head, value), value)) { ... }. The former approach does not, so you would have to do
push(&head, value);
if(contains(head, value)) { ... }
I would recommend using the indirect approach regardless because there are very few instances that you would want to check for containment after putting in a value.
This comparison:
if ((current -> val) == value)
it's comparing pointers. If you call your contains() function this way...
...
struct N val_to_find;
...
result = contains (list, &val_to_find);
You will never find the value, even if the contents of val_to_find are the same as the contents of any struct whose pointer is stored in the list.
If your intention for contains() is to find nodes that have the same data, and not just the same pointers, I'd suggest you something like this:
if (struct_n_comparing_function (current -> val, value) == EQUAL) ...
Where struct_n_comparing_function should have the following prototype:
int struct_n_comparing_function (struct N *a, struct N *b);
which compares the contents of the two structs pointed by a and b and return EQUAL if all the fields of the struct pointed by a have the same value as the fields of struct pointed by b.
So I am trying to learn how to create a binary tree in C so far I have got this.
void addRecordsToTree(struct date *in, struct date *root) {
if (root == NULL) {
root = malloc(sizeof(struct date));
root = in;
return;
} else {
//Right side of tree processing
if (compareTwoRecords(in, root) >= 0) {
addRecordsToTree(in, root->right);
return;
} else {
root->right = in;
return;
}
//Left side of tree processing.
if (compareTwoRecords(in, root) < 0) {
addRecordsToTree(in, root->left);
return;
} else {
root->left = in;
return;
}
}
}
int main() {
loadFiles();
struct date treeRoot;
struct date *old = malloc(sizeof(struct date));
old = loadContentsIntoHeap(files[file2014]);
addRecordsToTree(&old[0], &treeRoot);
addRecordsToTree(&old[1], &treeRoot);
addRecordsToTree(&old[2], &treeRoot);
addRecordsToTree(&old[3], &treeRoot);
addRecordsToTree(&old[4], &treeRoot);
addRecordsToTree(&old[5], &treeRoot);
printRecord(7, old);
return 0;
}
The problem is when I check the state of the program in a debugger there is just jumbled up data. I think it could be a type problem somewhere, I find pointers are bit of a mind boggling concept. Im not sure if I have used them right. So here is a screen shot of the debugger.
As you can see at the bottom struct called 'old' is the data I am trying to make the tree out of and treeRoot is where I am trying to place it but I can't understand why I get these garbage values.
Also what is up with the memory address of left and right? am I not creating them correctly.
Another observation I made is when I watch my code in the debugger it seems that root is never == NULL and never gets set, why?
You just did the following:
int x = 2;
int y = x;
y = 5;
Is the second line here necessary or the third one. It is a totally illogical program if you did this. You just did the same thing with a pointer instead of integer. You firstly had a pointer to the base address of dynamic memory then you just overwrote it by initializing it the second time.
And, the iterative approach is far better in comparison to the recursive one. I share the code for inserting a node in a binary tree both recursively and iteratively:
void insert(struct node *temp, struct node **root)
{
while (*root != NULL)
root = (*root)->element < temp->element ? &(*root)->left : &(*root)->right;
*root = temp;
}
#if 0
/* Recursive approach */
void insert(struct node *temp, struct node **root)
{
if(*root == NULL)
*root = temp;
else if ((*root)->element < temp->element)
insert(temp, &(*root)->left);
else
insert(temp, &(*root)->right);
}
#endif
void create_node(int x, struct node **root)
{
struct node *temp = (struct node *) malloc(sizeof(struct node));
if (temp == NULL)
printf("Unable to allocate memory. Free some space.\n");
else
{
temp->left = NULL;
temp->right = NULL;
temp->element = x;
insert(temp, root);
}
}
int main()
{
struct node *root = NULL;
create_node(1, &root);
create_node(2, &root);
create_node(3, &root);
return 0;
}
I saw an additional Problem in your "addRecordsToTree"-function:
the IF-block of the
"//Right side of tree processing"
will allways return from the function. regardless wether the "IF"-Expression is true or false.
So your left-leaves of thew tree will never be inserted. So you probalby should check/debug that function.
Disclaimer: This is for an assignment. I am not asking for explicit code answers, only help understanding why my code isn't working.
I am trying to implement a basic Binary Search Tree, but I am having problems with my _addNode(...) function.
Here's the problem. When I walk through my code with the debugger, I notice that leaf nodes are created infinitely on both sides (left and right) so aside from the creation of the root, there is never any point when a leaf node is NULL. The problem is that I am asking my program to create a new node whenever it finds a NULL value where a leaf would be. Therefore, if there are never any NULL values, there will never be any new leaves created, right?
The other issue I'm running into is with my compare(...) function. Stepping through it in the debugger shows it to iterate through the function several times, never actually returning a value. When it returns to the calling function, it drops back into the compare(...) function and loops infinitely. Again, I don't know why this is happening considering I have valid return statements in each if statement.
Here is all the code you'll probably need. If I left something out, let me know and I'll post it.
struct Node {
TYPE val;
struct Node *left;
struct Node *right;
};
struct BSTree {
struct Node *root;
int cnt;
};
struct data {
int number;
char *name;
};
int compare(TYPE left, TYPE right)
{
assert(left != 0);
assert(right != 0);
struct data *leftData = (struct data *) left;
struct data *rightData = (struct data *) right;
if (leftData->number < rightData->number) {
return -1;
}
if (leftData->number > rightData->number) {
return 1;
} else return 0;
}
void addBSTree(struct BSTree *tree, TYPE val)
{
tree->root = _addNode(tree->root, val);
tree->cnt++;
}
struct Node *_addNode(struct Node *cur, TYPE val)
{
assert(val != 0);
if(cur == NULL) {
struct Node * newNode = malloc(sizeof(struct Node));
newNode->val = val;
return newNode;
}
if (compare(val, cur->val) == -1) {
//(val < cur->val)
cur->left = _addNode(cur->left, val);
} else cur->right = _addNode(cur->right, val);
return cur;
}
Edit: Adding the below function(s)
int main(int argc, char *argv[])
{
struct BSTree *tree = newBSTree();
/*Create value of the type of data that you want to store*/
struct data myData1;
struct data myData2;
struct data myData3;
struct data myData4;
myData1.number = 5;
myData1.name = "rooty";
myData2.number = 1;
myData2.name = "lefty";
myData3.number = 10;
myData3.name = "righty";
myData4.number = 3;
myData4.name = "righty";
/*add the values to BST*/
addBSTree(tree, &myData1);
addBSTree(tree, &myData2);
addBSTree(tree, &myData3);
addBSTree(tree, &myData4);
/*Print the entire tree*/
printTree(tree);
/*(( 1 ( 3 ) ) 5 ( 10 ))*/
return 1;
}
Maybe you could try setting right and left to NULL right after malloc:
struct Node * newNode = malloc(sizeof(struct Node));
newNode->left = NULL;
newNode->right = NULL;
Check this line here (or the corresponding for left):
cur->right = _addNode(cur->right, val);
If cur->right == 0, it's fine. But if cur->right != 0, the node that was sitting there will be replaced by the return value of _addNode, which ultimately is not a whole branch, but just one node.
I like to explicitly 0-out values in a struct after a malloc using memset(newNode, 0, sizeof(struct Node)). Others might disagree.