I am building an AVL tree, using this structure:
typedef struct node* nodep;
typedef struct node {
int postal_number;
int h; /* height */
nodep left, right, parent;
} Node;
Everything is working great: rotating, searching and inserting. The problem is with deleting a node. It deletes most of the node, except for few leaves after deleting most of the tree. My delete code (the problematic part) is:
int delete_postal(nodep* t, int in_postal){
nodep to_delete;
nodep* the_parent;
to_delete = search_postal((*t), in_postal);
the_parent = &(to_delete->parent);
/** if the node we want to delete is a leave **/
if (to_delete->right == NULL && to_delete->left == NULL) {
if (to_delete == (*the_parent)->right){ /* the node we want to delete is right son */
(*the_parent)->right = to_delete->right;
else { /* the node we want to delete is left son */
(*the_parent)->left = to_delete->left;
}
I made it so (*the_parent) will be the node himself in the tree, and when debugging it stepping into the if, it acts as if it did the deletion (making the child of the node the child of the node's father - I'm not dealing with malloc right now), but it just does not do it. The father keeps on pointing to the node that I would like to delete and not to NULL.
All the rest of the first deletion (nodes and leaves) works fine with this syntax.
Does somebody know what I am missing?
Related
I'm building a database for a project, every entry has an RB-Tree that contains its attributes, but with the function that I made I cannot save the changes of the trees. I create and link the new nodes but when I return the new nodes are lost
x is the attribute (relation) identifier, used to look inside the tree and see if the relation already exists or not. id_orig is the entry of the database which I have to ad the relation to. tree * radice is the root of the tree and tree * nil is the nil pointer (substitute to NULL).
Add_rel_list is the function used to add a new relation after adding it on a new tree node if there isn't one.
insert fixup is used to check that the RB-Tree is balanced and respects it's definition
this is the add attribute function
struct tree *insert(unsigned long int x, char id_orig[50], char nome_relazione[50], struct tree *radice, struct tree * nil){ /*devo creare un nuovo nodo dell'albero */
if(radice==nil){
//this part is used to create a new root of the tree
}
else{
struct tree *newleaf=malloc(sizeof(struct tree)); /*creating a new node*/
//struct rel_list *lista=malloc(sizeof(struct rel_list));
newleaf->codice_rel=x; /*assign the node number*/
//newleaf->rel=lista; //link the list
newleaf->nome_rel=(char*)malloc(strlen(nome_relazione)+1);
strcpy(newleaf->nome_rel, nome_relazione);
newleaf->rel=add_rel_list(newleaf->rel, id_orig); /*helper function to add the list names*/
newleaf->num_rel=1; /*metto a 1 la quantit‡ del tipo di relazioni esistenti*/
struct tree *y= nil;
struct tree *w= radice;
while(w!=nil){
y=w;
if(newleaf->codice_rel<w->codice_rel){
w=w->left;
}
else{
w=w->right;
}
newleaf->prev=y;
if (y==nil){
radice=newleaf; //l'albero Ë vuoto
}
else if(newleaf->codice_rel<y->codice_rel){
if(newleaf->codice_rel<y->left->codice_rel){
newleaf->right=y->left;
}
else{
newleaf->left=y->left;
}
y->left=newleaf;
}
else{
if(newleaf->codice_rel<y->right->codice_rel){
newleaf->right=y->right;
}
else{
newleaf->left=y->right;
}
y->right=newleaf;
}
newleaf->black=0;
insert_fixup(radice, newleaf, nil);
return radice;
}
}
}
struct tree *insert_rel(unsigned long int x, char *id_orig, struct tree *radice, struct tree *nil, char *id_dest, char *nomerel, controllo * controllore){
//non relevant to the problem
return radice;
}
void insert_fixup(struct tree *radice, struct tree* nuovo, struct tree *nil){
//this function is used to balance the tree
}
the function should create a new node and link it inside the tree. the calls that are in the function and that I did not post are irrelevant to the problem, leftrotate and rightrotate are for the manipulation of the RB-Tree. when, later on, I look inside the tree for the attributes (relations) the only one saved is the first one, the root of the tree. using gdb the linking is correct but when I return to the main part of the code I lose all the new nodes of the tree
I have to do a function where, given an int n and a binary search tree, i have to convert the level n of the bst int a linked list.
for example if given the numer 2 and this tree
2
/ \
5 3
i have to make a liked list with 5 - 3
I have problems getting to the level given, and then getting each node on that level, because if i do reach the level i don't know how to reach the next node. Meaning, i can only get to the level on only one branch, and i cant think of any way to do it recursively.
so this is the struct for the bst and the linked chain:
struct nodo {
info_t dato;
nodo *anterior;
nodo *siguiente;
};
struct rep_cadena {
nodo *inicio;
nodo *final;
};
struct rep_binario {
info_t dato;
rep_binario *izq;
rep_binario *der;
};
and this is the function i cant figure out:
cadena_t nivel_en_binario(nat l, binario_t b)
i have tried using another function i already made, that calculates the height of a tree, but i can't stop on the wanted level.
nat altura_binario(binario_t b) {
if (b==NULL) return 0;
else return maximo(altura_binario(b->izq), altura_binario(b->der))+ 1;
}
where maximo() returns the highest number between the two given numbers.
You can do this by implementing Breadth-first_search algorithm, and modifying a little bit. Instead of enqueuing just nodes, you can enqueue pairs of (node, level) (where node level = parent level + 1), and then when enqueuing you can check if you have reached the desired level, and just output it as the result instead of queuing it further.
Pseudocode sketch:
target_level = ...read from input...
let level_nodes = ...an empty list...
let queue = ...an empty queue...
queue.enqueue((root_node, 0))
while queue is not empty:
node, level = queue.dequeue()
if level == target_level:
level_nodes.append(node)
else:
if node has left child:
queue.enqueue((left_child_node, level + 1))
if node has right child:
queue.enqueue((right_child_node, level + 1))
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.
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 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.