AVL-tree clearing - c++

I don't know, why _clear don't remove elements from a tree. Probably, needs to override destructor.
class Node {
T _val;
Node *_left;
Node *_right;
int _height;
explicit Node(T _val = T()) : _val(_val), _left(nullptr), _right(nullptr), _height(1) {};
friend class AVL_Tree<T>;
};
void AVL_Tree<T>::_clear(Node<T>*vertex) {
if(vertex != nullptr) {
_clear(vertex->_left);
_clear(vertex->_right);
delete vertex;
}
vertex = nullptr;
}
There are stay NULL linked nodes, why they don't delete?

The problem is here:
AVL_Tree<T>::_clear(Node<T>*vertex)
You passed a pointer to Node by value, but you want to change the pointer and have the change propagate out of the function. The C++ way to do this is pass a reference to the pointer-to-Node:
AVL_Tree<T>::_clear(Node<T>*&vertex)
PS. Don't begin an identifier with an underscore. That's reserved for library implementers. So:
AVL_Tree<T>::clear(Node<T>*&vertex)

Related

Modifying private pointer of object within same type (but different object) public method

I've been attempting to create a node class which mimics a node on a graph. Currently, storage of the predecessor and successor nodes are stored via a node pointer vector: std::vector<Node*> previous. The vectors for the predecessor/successor nodes are private variables and are accessible via setters/getters.
Currently, I am dealing with updating the pointer values when adding a new node. My current method to update the predecessor/successor nodes is through this method (the method is the same for successor/previous nodes, just name changes):
void set_next(std::vector<Node*> new_next) {
this->next.clear();
for (Node* node : new_next) {
this->next.push_back(node);
}
}
This works for the current node but I was wondering the best way to update the new_next nodes that are passed in, especially the most 'C++'-esque way to accomplish this. I have previously written a method which adds individual nodes to the successor/predecessor vector:
void add_next(Node* new_node, bool one_way = false) {
this->next.pushback(new_node);
if (!one_way) {
new_node->add_prev(this, one_way = true);
}
}
The one_way variable was used to determine the depth(?) of the add_next() call. If it is true, it will add to the successor set and then add to the predecessor set of new_node. Since the boolean value is set to false when the new_node->add_next() method call occurs, it will only add to the predecessor set and not attempt to call add_next()/add_prev() again. This solution does work, but I'd rather not have the one_way variable and would prefer that the method would be private.
Here is the structure of the class currently:
class Node {
private:
std::vector<Node*> previous;
std::vector<Node*> next;
boost::any data;
public:
std::vector<Node*> get_previous()
void set_previous(std::vector<Node*> new_previous)
std::vector<Node*> get_next()
void set_next(std::vector<Node*> new_next)
void add_prev(Node* new_node, bool one_way = false)
void add_next(Node* new_node, bool one_way = false)
}
Avoiding the one_way parameter, it seems my best solution would be to just create an add_next/add_prev method that only updates the current node, not the passed node, unlike the solution above. With this solution, when adding new nodes, I could call the inverse of the add_next/add_prev on the new node. However, I have an inkling there may be a better solution for this.
Thank you!
I think this should get you going (edge-cases left to you to figure out, if any):
template<typename T>
class Node {
// Everything made public for debugging purposes, change this to fit your needs
public:
std::vector<Node<T>*> previous;
std::vector<Node<T>*> next;
T data;
Node(T val) {
data = val;
}
void set_next(std::vector<Node<T>*>& new_next);
};
template<typename T>
void Node<T>::set_next(std::vector<Node<T>*>& new_next) {
next = new_next;
for (Node<T>* node : new_next)
node->previous.push_back(this);
}
int main() {
// Little proof of concept where 0 is parent to 1, 2, 3
Node<int> one = 1;
Node<int> two = 2;
Node<int> three = 3;
Node<int> zero = 0;
std::vector<Node<int>*> new_next = { &one , &two, &three };
zero.set_next(new_next);
return 0;
}
To sum up the differences:
Use templates, boost::any is non standard and is all around terrible for this task.
Leverage operators (= makes a copy of std::vectors).
Leverage reference types (this way you can modify the argument passed to your function).

Circular double linked list with smart pointers in c++

Is it possible to create a circular doubly-linked list using smart pointers in C++
struct Node {
int val;
shared_ptr<Node> next;
weak_ptr prev;
};
shared_ptr<Node> head;
But this will have a circular reference of shared pointers and thus not deallocate correctly.
Make the circular linked list a class itself (with whatever operations you need to build it, like append). Have its destructor break the link by setting tail->next = nullptr. It should not matter which link you break, so if you're not using a head and tail, just set any one of them nullptr, and you're good.
In my testing, I made a circular linked list, and the nodes did not destruct. Then at the end, I added tail->next = nullptr before it exited, and all the destructors fired correctly.
My original posted answer was rather light on details. This one gives a proper explanation of how you can achieve a circular linked list without a memory leak and still adhere to the Rule of Zero. The answer is basically the same, using a sentinel, but the mechanism is a little more involved than I had originally let on.
The trick is to use a sentinel type that behaves just like a list node, but in fact does not really have a shared pointer to the head of the list. To achieve this, the node class should be separated into a behavior object and a state object.
class NodeState {
std::shared_ptr<Node> next_;
std::weak_ptr<Node> prev_;
int value_;
NodeState (int v) : value_(v) {}
NodeState (std::shared_ptr<Node> p) : next_(p), prev_(p) {}
//...
};
class Node {
virtual ~Node () = default;
virtual NodeState & state () = 0;
std::shared_ptr<Node> & next () { return state().next_; }
std::weak_ptr<Node> & prev () { return state().prev_; }
int & value () { return state().value_; }
void insert (const std::shared_ptr<Node> &p) {
//...
}
};
Now, you can define a node implementation and a sentinel implementation.
class NodeImplementation : public Node {
NodeState state_;
NodeState & state () { return state_; }
NodeImplementation (int v) : state_(v) {}
//...
};
class NodeSentinel : public Node {
List &list_;
NodeSentinel (List &l) : list_(l) {}
NodeState & state () { return list_.sentinel_state_; }
};
The list itself contains a NodeState used by the sentinel object. Upon initialization, the list creates a sentinel object and initializes its state.
class List {
//...
NodeState sentinel_state_;
std::shared_ptr<Node> head () { return sentinel_state_.next_; }
std::shared_ptr<Node> sentinel () {
return std::shared_ptr<Node>(head()->prev());
}
//...
public:
List () : sentinel_state_(std::make_shared<NodeSentinel>(*this)) {}
//...
void push_front (int value) {
head()->insert(std::make_shared<NodeImplementation>(value));
}
void push_back (int value) {
sentinel()->insert(std::make_shared<NodeImplementation>(value));
}
//...
};
So, what does this organization do? It avoids the issue of a circular reference by using a sentinel node to act as the break. While the tail of the list points to the sentinel object, the sentinel object itself does not point to anything. Instead, it uses the state of the list itself to determine its next and previous neighbors.
Thus, the circular shared pointers only persists as long as the list exists. Once the list is destroyed, Item A loses its reference, and via the domino effect, Sentinel itself will be destroyed.
A fundamental point is that the sentinel object itself must never be exposed to the user of the list interface directly. It should remain internal to the list object at all times. It essentially represents end() in an STL like container, and logically, it can never be removed from the list (until the list itself is destroyed). In practice, this means removal operations on the list need to exit early if the passed in iterator represents the sentinel.
Demo
Try It Online
It is also possible to define a member function next() which can select between a shared or weak pointer.
#include <iostream>
#include <memory>
using namespace std;
struct T {
int n_;
shared_ptr<T> next_;
weak_ptr<T> weaknext_;
T(shared_ptr<T> next, int n) : next_(next), n_(n) {};
auto next() {
if (next_ == nullptr)
return shared_ptr<T>(weaknext_);
return next_;
}
~T() { cout << n_ << "ok\n"; }
};
int main() {
auto p0 = make_shared<T>(nullptr, 1);
auto p1 = make_shared<T>(p0, 2);
auto p2 = make_shared<T>(p1, 3);
p0->weaknext_ = p2; //makes the list circular
auto p = p2;
for (int i = 0; i < 5; ++i) {
cout << p->n_ << "\n";
p = p->next();
}
}

How to return pointer from linked list (in Class Template)

I've stumbled upon a problem with my linked list class.
I've one abstract class Shape and multiple classes inheriting from it, like Square or Triangle etc.
I'm storing them in my List class but I don't know how to return stored object back to the pointer of Shape.
Since my explanation may seem pretty vague here is some code with expected behaviour explained.
class Shape // abstract class
{
public:
int a;
//some member virtual methods
};
class Square : public Shape
{
//using the virtual methods from Shape
};
In my main file, this is how I want to use it:
int main()
{
List<Shape*> ShapeList;
Shape *ptr;
Square a(2, 1, 1); // size, x, y coordinates
ShapeList.add(ptr);
//up to this point everything works well
// now I want my list to return a pointer to it's member
// so I can modify it
Shape *listptr;
listptr = ShapeList.findInstanceAt(0); // here's my error
listptr->a = 5; // what I want to do next
}
So as you can see I'm havingtroubles with returning proper value from my list and I don't know how to solve this.
Here's my simplified list implementation:
template <class T> class Node
{
T data;
Node *next;
public:
inline T getData()
{
return data;
}
inline Node* getNext()
{
return next;
}
};
template <class T> class List
{
Node<T> *head, *tail;
public:
List() : head(NULL), tail(NULL) { }
T* findInstanceAt(int _k)
{
if (NULL == head)
{
cout << "\nList is empty.";
return NULL;
}
else
{
Node<T> *temp = new Node<T>;
temp = head;
for (size_t k = 0; k < _k; ++k)
{
if (NULL != temp->getNext()) temp = temp->getNext();
else return NULL;
}
return temp->getData;
}
}
}
Thanks in advance for any suggestions on how to make this work.
#EDIT
Ahh I forgot to add compiler errors that I'm getting:
Error 1 error C2440: '=' : cannot convert from 'Shape **' to 'Shape *'
Do you want to store Shapes or pointers to Shapes in the list? And do you want the findInstanceAt to return the node in the list or a pointer to the node in the list? At the moment you are not consistent on these things
You store Shape* nodes in the list but the findInstanceAt returns a pointer to the node - which is a Shape** object. This is what the compiler is complaining about
You probaly need to chang
T* findInstanceAt(int _k)
to
T findInstanceAt(int _k)

C++ problems with conversion

I have one semestral work (own double linked list) and our teacher want this definition of class DoubleList:
template <typename T> //just part of all methods
class DoubleList {
public:
DoubleList(void); //We HAVE TO follow this definitions
void AddFirst(const T &); //const!
T &AccessActual(void);
T RemoveFirst(void);
}
My question is, how can I define a node? AddFirst have const argument and other methods haven't. Data must be set in constructor and then they can't be changed. Is this task so limited or are here other ways to complete the task?
Here is my actual Node:
template <class U>
class Node{
Node<U> * next;
Node<U> * previous;
const U * data;
public:
Node(const U *data){ //
next = NULL;
previous = NULL;
this->data = data;
}
void SetNext(Node<U> *next) {
this->next = next;
}
Node<U> *GetNext(){ return next; }
void SetPrevious(Node<U> *previous) {
this->previous = previous;
}
Node<U> *GetPrevious(){ return previous; }
const U *GetData() { return data; }
};
In containers, it's usually better to have a copy of the data so change const U * data; to U data;
The Node constructor would be easier to use if it had this signature Node(const U& data). No pointers.
The GetData would also have to change. Return a reference. U& GetData().
It is dangerous to hold addresses of data items. If the user of the lists wants that functionality he can use a list that stored pointers (e.g. U=int*)
Your node class seems fine, although i would keep using template argument T instead of U, right now it is confusing.
Your AddFirst() method should simply create a new node and assign the correct next pointer to the new node and the correct prev pointer to the "old" first node and adjust the actual object? what does that refer to?
Your interface of nodes differs from this one returning a reference instead of a pointer. I find it quite strange that the AccessActual can always return an object, while when the list is empty this can be a nullptr??
example implementation:
void AddFirst(const T &)
{
Node<T>* newNode = new Node<T>(T);
Node<T>* current = &AccessActual(); // how can there be an actual when the list can be empty or is that impossible?
{
while( current.GetPrev() != nullptr )
{
current = *current.GetPrev();
}
current.SetPrev(newnode);
newnode->SetNext(current);
}
}

Converting to template class (NODES)

template <class Type>
class Node
{
public:
Node ()
{
}
Node (Type x, Node* nd)
{
data = x;
next = nd;
}
Node (Type x)
{
data = x;
next = NULL;
}
~Node (void)
{
}
Node (const Node* & nd)
{
data = nd->data;
next = nd->next;
}
Node & Node::operator = (const Node* & nd)
{
data = nd->data;
next = nd->next;
}
T data;
Node* next;
};
Do I replace every Node* with
Node*<Type>
I tried replacing it and tried running something like
Node* temp = myq.head;
but it says argument list for class template "Node" is missing. I'm not really sure how to work with Templates when I need the Node class itself being part of it
Every declaration of Node will need a type in <>.
For
Node* temp = myq.head;
it depends on what myq.head is defined as. If it's defined as Node<int>* then temp also has to be defined as Node<int>* temp. You always have to have the <> with template objects.
If you wanted to have Node* without knowing the type, you could use inheritance. Have a templated TypedNode class that inherits from a non-template Node class. You would be able to pass all those TypeNode<> objects around with Node*, but you wouldn't be able to get the value of the nodes back out without knowing their type.
I don't recommend this but If you really want to make nodelists with mixed types you'll need to track the types by either
Include an enum type in the base class that defines the type stored in the node, and define typedNode for each class, setting the enum in it's constructor, or returning it from a virtual method.
RTTI, Run Time Type Information http://en.wikipedia.org/wiki/Run-time_type_information