C recursively build tree using structure pointer - c

I'm now implementing Barnes-Hut Algorithms for simulating N-body problem. I only want to ask about the building-tree part.
There are two functions I made to build the tree for it.
I recursively build the tree, and print the data of each node while building and everything seems correct, but when the program is back to the main function only the root of the tree and the child of the root stores the value. Other nodes' values are not stored, which is weird since I printed them during the recursion and they should have been stored.
Here's some part of the code with modification, which I thought where the problem might be in:
#include<...>
typedef struct node{
int data;
struct node *child1,*child2;
}Node;
Node root; // a global variable
int main(){
.
set_root_and_build(); // is called not only once cuz it's actually in a loop
traverse(&root);
.
}
Here's the function set_root_and_build():
I've set the child pointers to NULL, but didn't show it at first.
void set_root_and_build(){
root.data = ...;
..// set child1 and child2 =NULL;
build(&root,...); // ... part are values of data for it's child
}
And build:
void build(Node *n,...){
Node *new1, *new2 ;
new1 = (Node*)malloc(sizeof(Node));
new2 = (Node*)malloc(sizeof(Node));
... // (set data of new1 and new2 **,also their children are set NULL**)
if(some condition holds for child1){ // else no link, so n->child1 should be NULL
build(new1,...);
n->child1 = new1;
//for debugging, print data of n->child1 & and->child2
}
if(some condition holds for child2){ // else no link, so n->child2 should be NULL
build(new2,...);
n->child1 = new2;
//for debugging, print data of n->child1 & and->child2
}
}
Nodes in the tree may have 1~2 children, not all have 2 children here.
The program prints out the correct data when it's in build() function recursion, but when it is back to main function and calls traverse(), it fails due to a segmentation fault.
I tried to print everything in traverse() and found that only the root, and root.child1, root.child2 stores the value just as what I've mentioned.
Since I have to called build() several times, and even in parallel, new1 and new2 can't be defined as global variables. (but I don't think they cause the problem here).
Does anyone know where it goes wrong?
The traverse part with debugging info:
void traverse(Node n){
...//print out data of n
if(n.child1!=NULL)
traverse(*(n.child1))
...//same for child2
}

You may not be properly setting the children of n when the condition does not hold. You might want this instead:
void set_root_and_build()
{
root.data = ...;
build(&root,...); // ... part are values of data for it's child
}
void build(Node *n,...)
{
n->child1 = n->child2 = NULL;
Node *new1, *new2;
new1 = (Node*) malloc(sizeof(Node));
new2 = (Node*) malloc(sizeof(Node));
// set data of new1 and new2 somehow (read from stdin?)
if (some condition holds for new1)
{
n->child1 = new1;
build(n->child1,...);
//for debugging, print data of n->child1
}
else
free(new1); // or whatever else you need to do to reclaim new1
if (some condition holds for new2)
{
n->child2 = new2;
build(n->child2,...);
//for debugging, print data of n->child2
}
else
free(new2); // or whatever else you need to do to reclaim new2
}
Of course, you should be checking the return values of malloc() and handling errors too.
Also, your traversal is a bit strange as it recurses by copy rather than reference. Do you have a good reason for doing that? If not, then maybe you want:
void traverse(Node *n)
{
...//print out data of n
if (n->child1 != NULL)
traverse(n->child1)
...//same for child2
}

The problem in your tree traversal is that you certainly process the tree until you find a node pointer which is NULL.
Unfortunately when you create the nodes, these are not initialized neither with malloc() nor with new (it would be initialized with calloc() but this practice in cpp code is as bad as malloc()). So your traversal continues to loop/recurse in the neverland of random pointers.
I propose you to take benefit of cpp and change slightly your structure to:
struct Node { // that's C++: no need for typedef
int data;
struct node *child1,*child2;
Node() : data(0), child1(nullptr), child2(nullptr) {} // Makes sure that every created are first initalized
};
And later get rid of your old mallocs. And structure the code to avoid unnecessary allocations:
if(some condition holds for child1){ // else no link, so n->child1 should be NULL
new1=new Node; // if you init it here, no need to free in an else !!
build(new1,...);
n->child1 = new1;
...
}
if (... child2) { ... }
Be aware however that poitners allocated with new should be released with delete and note with free().
Edit: There is a mismatch in your code snippet:
traverse(&root); // you send here a Node*
void traverse(Node n){ // but your function defines an argument by value !
...
}
Check that you didn't overllok some warnings from the compiler, and that you have no abusive cast in your code.

Related

How to handle malloc failing and returning NULL?

I'm a bit confused on how to check if a memory allocation failed in order to prevent any undefined behaviours caused by a dereferenced NULL pointer.
I know that malloc (and similiar functions) can fail and return NULL, and that for this reason the address returned should always be checked before proceeding with the rest of the program. What I don't get is what's the best way to handle these kind of cases. In other words: what is a program supposed to do when a malloc call returns NULL?
I was working on this implementation of a doubly linked list when this doubt raised.
struct ListNode {
struct ListNode* previous;
struct ListNode* next;
void* object;
};
struct ListNode* newListNode(void* object) {
struct ListNode* self = malloc(sizeof(*self));
if(self != NULL) {
self->next = NULL;
self->previous = NULL;
self->object = object;
}
return self;
}
The initialization of a node happens only if its pointer was correctly allocated. If this didn't happen, this constructor function returns NULL.
I've also written a function that creates a new node (calling the newListNode function) starting from an already existing node and then returns it.
struct ListNode* createNextNode(struct ListNode* self, void* object) {
struct ListNode* newNext = newListNode(object);
if(newNext != NULL) {
newNext->previous = self;
struct ListNode* oldNext = self->next;
self->next = newNext;
if(oldNext != NULL) {
newNext->next = oldNext;
oldNext->previous = self->next;
}
}
return newNext;
}
If newListNode returns NULL, createNextNode as well returns NULL and the node passed to the function doesn't get touched.
Then the ListNode struct is used to implement the actual linked list.
struct LinkedList {
struct ListNode* first;
struct ListNode* last;
unsigned int length;
};
_Bool addToLinkedList(struct LinkedList* self, void* object) {
struct ListNode* newNode;
if(self->length == 0) {
newNode = newListNode(object);
self->first = newNode;
}
else {
newNode = createNextNode(self->last, object);
}
if(newNode != NULL) {
self->last = newNode;
self->length++;
}
return newNode != NULL;
}
if the creation of a new node fails, the addToLinkedList function returns 0 and the linked list itself is left untouched.
Finally, let's consider this last function which adds all the elements of a linked list to another linked list.
void addAllToLinkedList(struct LinkedList* self, const struct LinkedList* other) {
struct ListNode* node = other->first;
while(node != NULL) {
addToLinkedList(self, node->object);
node = node->next;
}
}
How should I handle the possibility that addToLinkedList might return 0? For what I've gathered, malloc fails when its no longer possible to allocate memory, so I assume that subsequent calls after an allocation failure would fail as well, am I right? So, if 0 is returned, should the loop immediately stop since it won't be possible to add any new elements to the list anyway?
Also, is it correct to stack all of these checks one over another the way I did it? Isn't it redundant? Would it be wrong to just immediately terminate the program as soon as malloc fails? I read that it would be problematic for multi-threaded programs and also that in some istances a program might be able to continue to run without any further allocation of memory, so it would be wrong to treat this as a fatal error in any possible case. Is this right?
Sorry for the really long post and thank you for your help!
It depends on the broader circumstances. For some programs, simply aborting is the right thing to do.
For some applications, the right thing to do is to shrink caches and try the malloc again. For some multithreaded programs, just waiting (to give other threads a chance to free memory) and retrying will work.
For applications that need to be highly reliable, you need an application level solution. One solution that I've used and battle tested is this:
Have an emergency pool of memory allocated at startup.
If malloc fails, free some of the emergency pool.
For calls that can't sanely handle a NULL response, sleep and retry.
Have a service thread that tries to refill the emergency pool.
Have code that uses caching respond to a non-full emergency pool by reducing memory consumption.
If you have the ability to shed load, for example, by shifting load to other instances, do so if the emergency pool isn't full.
For discretionary actions that require allocating a lot of memory, check the level of the emergency pool and don't do the action if it's not full or close to it.
If the emergency pool gets empty, abort.
How to handle malloc failing and returning NULL?
Consider if the code is a set of helper functions/library or application.
The decision to terminate is best handled by higher level code.
Example: Aside from exit(), abort() and friends, the Standard C library does not exit.
Likewise returning error codes/values is a reasonable solution for OP's low-level function sets too. Even for addAllToLinkedList(), I'd consider propagating the error in the return code. (Non-zero is some error.)
// void addAllToLinkedList(struct LinkedList* self, const struct LinkedList* other) {
int addAllToLinkedList(struct LinkedList* self, const struct LinkedList* other) {
...
if (addToLinkedList(self, node->object) == NULL) {
// Do some house-keepeing (undo prior allocations)
return -1;
}
For the higher level application, follow your design. For now, it may be a simple enough to exit with a failure message.
if (addAllToLinkedList(self, ptrs)) {
fprintf(stderr, "Linked List failure in %s %u\n", __func__, __LINE__);
exit(EXIT_FAILURE);
}
Example of not exiting:
Consider a routine that read a file into a data structure with many uses of LinkedList and the file was somehow corrupted leading to excessive memory allocations. Code may want to simply free everything for that file (but just for that file), and simply report to the user "invalid file/out-of-memory" - and continue running.
if (addAllToLinkedList(self, ptrs)) {
free_file_to_struct_resouces(handle);
return oops;
}
...
return success;
Take away
Low level routines indicate an error somehow. Higher level routines can exit code if desired.

How to insert nodes in tree in C from right to left?

Now, I understand that code below works only for root and its children, but I don't know how to expand it. Every node must have children before passing on "grandchildren". Thank you.
void insert_node(IndexTree **root, Node *node) {
IndexTree *temp = (IndexTree*)malloc(sizeof(IndexTree));
memcpy(&temp->value.cs, node, sizeof(Node));
temp->left = NULL;
temp->right = NULL;
temp->tip=1;
if ((*root) == NULL) {
*root = temp;
(*root)->left = NULL;
(*root)->right = NULL;
}
else {
while (1) {
if ((*root)->right == NULL) {
(*root)->right = temp;
break;
}
else if ((*root)->left == NULL) {
(*root)->left = temp;
break;
}
}
}
Use recursive functions.
Trees are recursive data types (https://en.wikipedia.org/wiki/Recursive_data_type). In them, every node is the root of its own tree. Trying to work with them using nested ifs and whiles is simply going to limit you on the depth of the tree.
Consider the following function: void print_tree(IndexTree* root).
An implementation that goes over all values of the trees does the following:
void print_tree(IndexTree* root)
{
if (root == NULL) return; // do NOT try to display a non-existent tree
print_tree(root->right);
printf("%d\n", root->tip);
print_tree(root->left);
}
The function calls itself, which is a perfectly legal move, in order to ensure that you can parse an (almost) arbitrarily deep tree. Beware, however, of infinite recursion! If your tree has cycles (and is therefore not a tree), or if you forget to include an exit condition, you will get an error called... a Stack Overflow! Your program will effectively try to add infinite function calls on the stack, which your OS will almost certainly dislike.
As for inserting, the solution itself is similar to that of printing the tree:
void insert_value(IndexTree* root, int v)
{
if (v > root->tip) {
if (root->right != NULL) {
insert_value(root->right, v);
} else {
// create node at root->right
}
} else {
// same as above except with root->left
}
}
It may be an interesting programming question to create a Complete Binary Tree using linked representation. Here Linked mean a non-array representation where left and right pointers(or references) are used to refer left and right children respectively. How to write an insert function that always adds a new node in the last level and at the leftmost available position?
To create a linked complete binary tree, we need to keep track of the nodes in a level order fashion such that the next node to be inserted lies in the leftmost position. A queue data structure can be used to keep track of the inserted nodes.
Following are steps to insert a new node in Complete Binary Tree. (Right sckewed)
1. If the tree is empty, initialize the root with new node.
2. Else, get the front node of the queue.
……. if the right child of this front node doesn’t exist, set the right child as the new node. //as per your case
…….else If the left child of this front node doesn’t exist, set the left child as the new node.
3. If the front node has both the left child and right child, Dequeue() it.
4. Enqueue() the new node.

Implementing a deletion function for a binary search tree in C

I've been trying to implement a function in C that deletes a node in a binary tree that should (theoretically) take care of three all cases, i.e.:
Node is a leaf
Node has one child
Node has two children
Is there a way to handle the whole deletion function without checking separately each case? As a commenter below noted I do check for a lot of cases and perhaps the whole problem can be addressed recursively by checking for one fundamental case.
I'm particularly interested in the case where I delete a node within the tree that has a parent and itself is a parent of two children nodes.
Both answers below have been useful but I don't think they address the problem in its entirety.
Here's what I have:
typedef struct Node
{
int key;
int data;
struct Node *left;
struct Node *right;
struct Node *parent;
} Node;
/* functions that take care of inserting and finding a node and also traversing and freeing the tree */
...
void delete(Node *root, int key)
{
Node *target = find(root, key); // find will return the node to be deleted
Node *parent = target->parent; // parent of node to be deleted
// no children
if (target->left == NULL && target->right == NULL)
{
// is it a right child
if (target->key > parent->key)
parent->right = NULL;
// must be a left child
else
parent->left = NULL;
free(target);
}
// one child
else if ((target->left == NULL && target->right != NULL) || (target->left != NULL && target->right == NULL))
{
// here we swap the target and the child of that target, then delete the target
Node *child = (target->left == NULL) ? target->right : target->left;
child->parent = parent;
if (parent->left == target) parent->left = child;
else if (parent->right == target) parent->right = child;
free(target);
}
// two children
else
{
// find the largest node in the left subtree, this will be the node
// that will take the place of the node to be deleted
Node *toBeRepl = max(target->left);
// assign the data of the second largest node
target->key = toBeRepl->key;
target->data = toBeRepl->data;
// if new node immediately to the left of target
if (toBeRepl == target->left)
{
target->left = toBeRepl->left;
Node *newLeft = target->left;
if (newLeft != NULL) newLeft->parent = target;
}
else
{
delete(target->left, toBeRepl->key);
// Node *replParent = toBeRepl->parent;
// replParent->right = NULL;
}
}
I would greatly appreciate your feedback.
edit: Just to clarify, I'm trying to delete a particular node without touching its subtrees (if there are any). They should remain intact (which I've handled by swapping the values of the node to be deleted and (depending on the case) one of the nodes of its substrees).
edit: I've used as a reference the following wikipedia article:
http://en.wikipedia.org/wiki/Binary_search_tree#Deletion
Which is where I got the idea for swapping the nodes values in case of two children, particularly the quote:
Call the node to be deleted N. Do not delete N. Instead, choose either
its in-order successor node or its in-order predecessor node, R.
Replace the value of N with the value of R, then delete R.
There is some interesting code in C++ there for the above case, however I'm not sure how exactly the swap happens:
else //2 children
{
temp = ptr->RightChild;
Node<T> *parent = nullptr;
while(temp->LeftChild!=nullptr)
{
parent = temp;
temp = temp->LeftChild;
}
ptr->data = temp->data;
if (parent!=nullptr)
Delete(temp,temp->data);
else
Delete(ptr->rightChild,ptr->RightChild->data);
}
Could somebody please explain what's going on in that section? I'm assuming that the recursion is of a similar approach as to the users comments' here.
I don't see any "inelegance" in the code, such formatting and commented code is hard to come by. But yes, you could reduce the if-else constructs in your delete function to just one case. If you look at the most abstract idea of what deletion is doing you'll notice all the cases basically boil down to just the last case (of deleting a node with two children).
You'll just have to add a few lines in it. Like after toBeRepl = max(left-sub-tree), check if it's NULL and if it is then toBeRepl = min(right-sub-tree).
So, Case 1 (No children): Assuming your max() method is correctly implemented, it'll return NULL as the rightmost element on the left sub-tree, so will min() on the right sub-tree. Replace your target with the toBeRepl, and you'll have deleted your node.
Case 2 (One child): If max() does return NULL, min() won't, or vice-versa. So you'll have a non-NULL toBeRepl. Again replace your target with this new toBeRepl, and you're done.
Case 3 (Two children): Same as Case 2, only you can be sure max() won't return NULL.
Therefore your entire delete() function would boil down to just the last else statement (with a few changes). Something on the lines of:
Node *toBeRepl = max(target->left);
if toBeRepl is NULL
{
toBeRepl = min(target->right);
}
if toBeRepl is not NULL
{
target->key = tobeRepl->key;
target->data = toBeRepl->data;
deallocate(toBeRepl); // deallocate would be a free(ptr) followed by setting ptr to NULL
}
else
{
deallocate(target);
}
I would do it using recursion, assuming that you have null at the end of your tree, finding null would be the 'go back' or return condition.
One possible algorithm would be:
Node* delete(Node *aNode){
if(aNode->right != NULL)
delete(aNode->right);
if(aNode->left != NULL)
delete(aNode->left);
//Here you're sure that the actual node is the last one
//So free it!
free(aNode);
//and, for the father to know that you're now empty, must return null
return NULL;
}
It has some bugs, for sure, but is the main idea.
This implementation is dfs like.
Hope this helps.
[EDIT] Node *aNode fixed. Forgot the star, my bad.
I finished this a long time ago and I thought it would be good to add a sample answer for people coming here with the same problem (considering the 400+ views this question has accumulated):
/* two children */
else
{
/* find the largest node in the left subtree (the source), this will be the node
* that will take the place of the node to be deleted */
Node* source = max(target->left);
/* assign the data of that node to the one we originally intended to delete */
target->key = source->key;
target->data = source->data;
/* delete the source */
delete(target->left, source->key);
}
Wikipedia has an excellent article that inspired this code.

How to use free on a handle inside a list?-> C -> windows API

I have a list in C that is something like this:
typedef struct _node
{
int number;
DWORD threadID;
HANDLE threadH;
struct *_node next;
} *node;
And you have somthing like this:
node new_node = malloc(sizeof(node));
As you may have guessed out, this list will store information for threads, including their handlers and Id's. Still I am having trouble when I try to do this:
free(new_node);
Everytime I try to do this I encounter an unexpected error, VS saying that there was a data corruption. I've pinned down as much as possible and I found that the problem resides when I try to use free the handle.
I've searched on MSDN how to do this but the only thing I can find is the function that closes the thread (which is not intended here, since I want the thread to run, just deleting it's record from the list).
The question is: how I am supposed to free an handle from the memory? (Considering that this is only a copy of the value of the handle, the active handle is not being deleted).
EDIT: This is the function to insert nodes from the list:
int insereVisitanteLista(node* lista, DWORD threadID, HANDLE threadH, int num_visitante)
{
node visitanteAnterior;
node novoVisitante = (node)malloc(sizeof(node));
if(novoVisitante == NULL)
return 0;
novoVisitante->threadID = threadID;
novoVisitante->threadH = threadH;
novoVisitante->number = num_visitante;
novoVisitante->next = NULL;
if(*lista == NULL)
{
*lista = novoVisitante;
return 1;
}
visitanteAnterior = *lista;
while(visitanteAnterior->next != NULL)
visitanteAnterior = visitanteAnterior->next;
visitanteAnterior->next =novoVisitante;
return 1;
}
And this is the function to delete nodes:
int removeVisitanteLista(node * lista, DWORD threadID)
{
node visitanteAnterior = NULL, visitanteActual;
if(*lista == NULL)
return 0;
visitanteActual = *lista;
if((*lista)->threadID == threadID)
{
*lista = visitanteActual->next;
visitanteActual->next = NULL;
free(visitanteActual);
return 1;
}
while(visitanteActual != NULL && visitanteActual->threadID != threadID)
{
visitanteAnterior = visitanteActual;
visitanteActual = visitanteActual->next;
}
if (visitanteActual == NULL)
return 0;
visitanteAnterior->next = visitanteActual->next;
free(visitanteActual);
return 1;
}
What exactly is a node that you are trying to free? Is this a pointer to a struct _node? If yes, have you allocated it previously? If no, free is not needed, otherwise you have to check if node is not NULL and make sure you do not free it multiple times. It is hard to guess what you are doing and where is an error without a minimal working example reproducing the problem. The only thing I can suggest is to read about memory management in C. This resource might help.
UPDATE:
node in your code is a pointer to _node. So sizeof (node) is a size of a pointer, which is either 4 or 8 bytes (depending on architecture). So you allocate 8 bytes, for example, but assume you have a pointer to the structure which is much larger. As a result, you corrupt memory, and behavior of the program becomes undefined. So changing node novoVisitante = (node)malloc(sizeof(node)) to node novoVisitante = (node)malloc(sizeof(_node)) should fix the problem.
You haven't shown us the context of your call to free() so I need to speculate a little but my first concern is that you didn't mention removing the node from the list before deleting it.
Start by unlinking the node by modifying the next field of the previous (or head) node. If you still get the error, then you have corrupted memory somehow by writing past the end of one of your allocated memory structures or something similar.
Also, I assume node is a pointer. You really haven't provided much information about what you're doing.

Traverse tree without recursion and stack in C

How to traverse each node of a tree efficiently without recursion in C (no C++)?
Suppose I have the following node structure of that tree:
struct Node
{
struct Node* next; /* sibling node linked list */
struct Node* parent; /* parent of current node */
struct Node* child; /* first child node */
}
It's not homework.
I prefer depth first.
I prefer no additional data struct needed (such as stack).
I prefer the most efficient way in term of speed (not space).
You can change or add the member of Node struct to store additional information.
If you don't want to have to store anything, and are OK with a depth-first search:
process = TRUE;
while(pNode != null) {
if(process) {
//stuff
}
if(pNode->child != null && process) {
pNode = pNode->child;
process = true;
} else if(pNode->next != null) {
pNode = pNode->next;
process = true;
} else {
pNode = pNode->parent;
process = false;
}
}
Will traverse the tree; process is to keep it from re-hitting parent nodes when it travels back up.
Generally you'll make use of a your own stack data structure which stores a list of nodes (or queue if you want a level order traversal).
You start by pushing any given starting node onto the stack. Then you enter your main loop which continues until the stack is empty. After you pop each node from the stack you push on its next and child nodes if not empty.
This looks like an exercise I did in Engineering school 25 years ago.
I think this is called the tree-envelope algorithm, since it plots the envelope of the tree.
I can't believe it is that simple. I must have made an oblivious mistake somewhere.
Any mistake regardless, I believe the enveloping strategy is correct.
If code is erroneous, just treat it as pseudo-code.
while current node exists{
go down all the way until a leaf is reached;
set current node = leaf node;
visit the node (do whatever needs to be done with the node);
get the next sibling to the current node;
if no node next to the current{
ascend the parentage trail until a higher parent has a next sibling;
}
set current node = found sibling node;
}
The code:
void traverse(Node* node){
while(node!=null){
while (node->child!=null){
node = node->child;
}
visit(node);
node = getNextParent(Node* node);
}
}
/* ascend until reaches a non-null uncle or
* grand-uncle or ... grand-grand...uncle
*/
Node* getNextParent(Node* node){
/* See if a next node exists
* Otherwise, find a parentage node
* that has a next node
*/
while(node->next==null){
node = node->parent;
/* parent node is null means
* tree traversal is completed
*/
if (node==null)
break;
}
node = node->next;
return node;
}
You can use the Pointer Reversal method. The downside is that you need to save some information inside the node, so it can't be used on a const data structure.
You'd have to store it in an iterable list. a basic list with indexes will work. Then you just go from 0 to end looking at the data.
If you want to avoid recursion you need to hold onto a reference of each object within the tree.

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