Creating BST from sorted array in C++ - c++

Can you help me understand why this code randomly results in memory access violation? Goal is to generate a binary search tree from sorted list.
Stepping through code I noticed a behavior where node pointer randomly changes when a recursive calls to fromSortedArray() function is returned. To give you a contex I am compiling this app using XCODE.
#include <iostream>
using namespace std;
class BST{
private:
struct Node {
int val;
struct Node *left, *right;
} *root;
public:
BST(){
this->root = NULL;
}
struct Node * fromSortedArray(int data[], int left, int right){
if(left>right) return NULL;
int m = left+(right -left)/2;
struct Node *node = (struct Node *) malloc(sizeof(struct Node*));
node->val = data[m];
node->left = this->fromSortedArray(data, left, m-1);
node->right = this->fromSortedArray(data, m+1, right);
return node;
}
void fromSortedArray(int data[], int n){
this->root = fromSortedArray(data, 0, n-1);
}
void deleteTree(struct Node *root){
if(root==NULL) return;
deleteTree(root->left);
deleteTree(root->right);
delete root;
}
void deleteTree(){
this->deleteTree(this->root);
}
void traverse(struct Node *root){
if(root == NULL) return;
if(root->left!=NULL)traverse(root->left);
printf("%d ", root->val);
if(root->right!=NULL)traverse(root->right);
}
void traverse(){
this->traverse(this->root);
}
~BST(){
deleteTree();
}
};
int main(int argc, char * argv[]){
BST tree;
int data[] = {2,3,5,6,7,9};
tree.fromSortedArray(data, 6);
tree.traverse();
cout << "\n";
return 0;
}

I think this line is wrong.
struct Node *node = (struct Node *) malloc(sizeof(struct Node *));
should be
sizeof(struct Node) becacuse you want to apply memory for Node not Node*

The Nodes are being malloced. One cannot combine malloc and delete. malloc must be matched with free and new must be matched with delete.
Avoid malloc. It gets you bytes of memory and nothing more, and you can easily miscalculate the number of bytes required. malloc should only be used in C++ in rare edge cases. More reading on the topic: In what cases do I use malloc vs new?
Use new instead of malloc and if at all possible, don't use new either. Use an Automatic variable, a library container, or a smart pointer instead.
This is a tree structure and, opinion here, so long as you observe the Rules of Three and Five you can get away with raw, stupid pointers because it's the the tree structure's job to handle the management of the nodes. That said, play around a bit with std::unique_ptr and std::make_unique when you get a chance. They can make your life easier.
Sticking with raw pointers for now, replace
struct Node *node = (struct Node *) malloc(sizeof(struct Node*));
with
Node *node = new Node;

Related

deep copy in struct pointers

in this code, I think it must do deep copy because I'm passing pointers, but it doesn't.
I think it must print 3, but print 0. what should I do to solve this? i want to have a deep copy instead of a shallow copy.
struct node{
int number = 0;
struct node* right_child = NULL;
struct node* left_child = NULL;
};
void test(struct node* node1 , struct node* node2){
node1 = node2;
}
int main(){
struct node* a1 = new struct node;
struct node* a2 = new struct node;
a2->number = 3;
test(a1 , a2);
cout << a1->number;
}
The simple C-ish way: We ad a function that recursively clones the nodes
#include <iostream>
struct node{
int number = 0;
node* right_child = nullptr; // nullptr safer than NULL
node* left_child = nullptr;
};
node * clone(const node * src){
if (src) { // there is a node. Copy it.
return new node{src->number,
clone(src->right_child), // clone right
clone(src->left_child)}; // clone left
}
else { // no node. Nothing to do and end the branch
return nullptr;
}
}
void test(node*& node1, // reference to pointer
node* node2){
delete node1; // Free storage currently used by node1
// But what of the nodes that could be in its tree?
// Think on this. Ask another question if necessary.
// this is where the simple way starts to fall down
node1 = clone(node2);
}
int main(){
struct node* a1 = new node; // wasted memory. We're about to replace it.
// need a more complicated tree to prove deep copy works
struct node* a2 = new node{3,
new node{4, nullptr, nullptr}, // a right node
nullptr}; // no left node
// a2->number = 3; done above now
test(a1 , a2);
std::cout << a1->number << ' '
<< a1->right_child->number;
delete a1; // clean up
delete a2;
}
More complicated canonical C++ way using The Rule Of Three and The Copy and Swap Idiom
#include <iostream>
#include <algorithm> // for std::swap
struct node{
int number = 0;
node* right_child = nullptr; // nullptr safer than NULL
node* left_child = nullptr;
static node * clone(node * src)
{
if (src)
{
return new node(*src);
}
return nullptr;
}
// generic constructor
node(int number = 0,
node* right_child = nullptr,
node* left_child = nullptr):
number(number),
right_child(right_child),
left_child(left_child)
{
}
//copy constructor
node (const node & src):
number(src.number),
right_child(clone(src.right_child)),
left_child(clone(src.left_child))
{
}
// assignment operator via copy and swap.
node & operator=(node src)
{
std::swap(number, src.number);
std::swap(right_child, src.right_child);
std::swap(left_child, src.left_child);
return *this;
}
// destructor
~node()
{
delete right_child;
delete left_child;
}
};
void test(node* node1,
node* node2){
*node1 = *node2; // now, thanks to all of the infrastructure above, we can
// assign nodes with the dumb old = operator. All we have to do
// is dereference the pointers.
}
int main(){
struct node* a1 = new node; // wasted memory. We're about to replace it.
// need a more complicated tree to prove deep copy works
struct node* a2 = new node{3,
new node{4, nullptr, nullptr}, // a right node
nullptr}; // no left node
// a2->number = 3; done above now
test(a1 , a2);
std::cout << a1->number << ' '
<< a1->right_child->number;
delete a1; // clean up
delete a2;
}
All of the nodes are now self-managing.
But, my opinion, the nodes should be kept as dumb as they are in the simple example. They shouldn't need to know about the tree at all. Knowledge of the tree should be in a Tree class that uses the nodes as simple containers. The Tree class should do all of the management of the nodes--creating, copying, cloning, deleting--because only it knows everything necessary to represent the tree. The users of the tree shouldn't even know nodes exist. They put data into the tree, take data out of the tree, and observe the tree without caring how the tree works and being able to so something stupid like
delete a_node;
and blow the crap out of data the tree isn't done with yet.
The Tree class preferably works iteratively rather than recursively so that the trees can be arbitrarily sized. Unless you're really careful recursing through a large tree to clone it, delete it, display it, or pretty much anything else you do with a tree runs the risk of exhausting automatic storage and causing what's typically called a Stack Overflow.
just use
void test(struct node *node1, struct node *node2) { *node1 = *node2; }
instead of
void test(struct node *node1, struct node *node2) { node1 = node2; }
and it will print 3.
This is because...
when you do node1 = node2; int the test1, you assign the pointer itself, not the structure pointed to by the pointer.When the function ends, the parameters node1 and node2 will be destroyed, so you have done nothing...

C++ delete node and its child nodes [duplicate]

This question already has answers here:
Binary Search Tree Destructor
(6 answers)
Closed 2 years ago.
Please help me. I am stuck at this.
What am I trying to do: Binary search tree.
I am a C# developer and I learn C++ for about 2 weeks, therefore don't be so harsh with me and that's why pointers are still difficult for me.
I have a struct Node
struct Node
{
int Value;
Node* _LeftNode;
Node* _RightNode;
Node(int value)
: Value(value), _LeftNode(NULL), _RightNode(NULL)
{
}
};
and a Delete() function in BinarySearchTree.cpp
void BinarySearchТрее::Delete(Node* node)
{
if (node)
{
Delete(node->_LeftNode);
Delete(node->_RightNode);
delete(node);
node = NULL;
}
}
I want to delete the node and all of its child nodes.
When I first step in the recursion... For example:
I have two child nodes with values 10 and 19.
With recursion, I delete the nodes and set the pointers to NULL.
And here is the problem:
When I came out from the recursion the nodes are not NULL, but something strange.
And this is my problem. Why when I am in the recursion and I NULL the pointer everything is fine, but when I come out the pointer is something else.
As I talked in the comments, I think the thing is that how we can reset the pointer of the parent's(left or right child) of the initially passed node. (recursively deleting a node and its all children looks good.)
And I don't think it is possible in your current design. As Node does not contain a pointer to its parent, so there is no way to know who's the parent. node = NULL sets just the argument(local variable)'s value so it is pointless.
The C++ way would be to use std::unique_ptr.
struct Node
{
int Value;
std::unique_ptr<Node> LeftNode;
std::unique_ptr<Node> RightNode;
Node(int value)
: Value(value)
{
}
};
Then to destroy a node and all of its children, you'd call reset on the appropriate std::unique_ptr<Node>
I think what you actually want ist this:
struct Node
{
int Value;
Node* _LeftNode;
Node* _RightNode;
Node(int value)
: Value(value), _LeftNode(NULL), _RightNode(NULL)
{
}
~Node() {
delete _LeftNode;
delete _RightNode;
}
};
This way you are using the destructor to clean up recursivly.
delete nullptr is ok btw.
EDIT:
the unique_ptr<> usage in one of the other answers is probably the smarter way to do this.
Given:
struct Node
{
int data = 0;
struct Node * left = nullptr, * right = nullptr;
Node(int data) { this->data = data; }
};
This recursive function deletes a node & its childs (+ one comment):
void DeleteTree(struct Node* node) // A copy of the caller pointer
{
if (node)
{
DeleteTree(node->left); // Recur on left subtree
DeleteTree(node->right); // Recur on right subtree
delete node;
// node = nullptr; <-- This line is useless
}
}
To your wondering "but when I come out the pointer is something else":
There is no point in node = nullptr line, since when you call DeleteTree(my_node) function, node is a copy of my_mode, so when you set node = nullptr it has no effect on my_node that on exit from DeleteTree(my_node) points to a deleted, invalid object.
--
Possible solution:
#define DELETE_TREE(node) DeleteTree(node); node = nullptr; // Macro
int main()
{
struct Node* root = new Node(1);
root->left = new Node(2);
root->right = new Node(3);
root->left->left = new Node(4);
root->left->right = new Node(5);
DELETE_TREE(root->left); // root->left became nullptr
DELETE_TREE(root); // root became nullptr
return 0;
}
After DeleteTree function, the caller pointer points to an invalid object since its object already released. A possible solution is to define a DELETE_TREE Macro to "auto-nullify" the caller pointer after DeleteTree function.
--
Implementation with Modern C++ Smart Pointers:
#include <memory>
struct Node
{
int data = 0;
std::unique_ptr<Node> left, right;
Node(int data) { this->data = data; }
};
int main()
{
std::unique_ptr<Node> root;
root = std::make_unique<Node>(1);
root->left = std::make_unique<Node>(2);
root->right = std::make_unique<Node>(3);
root->left->left = std::make_unique<Node>(4);
root->left->right = std::make_unique<Node>(5);
root.reset();
return 0;
}

How to assign an instance of an object to a pointer that was null?

Code:
#include <iostream>
using namespace std;
class Node {
public:
Node *next;
int value;
Node(int value) {
this->next = nullptr;
this->value = value;
}
};
class LinkedList {
private:
Node *head;
Node *tail;
public:
LinkedList() {
this->head = nullptr;
this->tail = nullptr;
}
void addToEnd(int value) {
if(head == nullptr)
this->head = new Node(value);
else
this->tail->next = new Node(value);
this->tail = this->tail->next;
}
void print() {
for(Node *n = this->head; n != nullptr; n = n->next)
cout<<n->value<<" ";
cout<<endl;
}
};
int main() {
LinkedList *list = new LinkedList();
list->addToEnd(21);
list->addToEnd(25);
list->addToEnd(56);
list->addToEnd(24);
list->print();
return 0;
}
My problem is, when I am assigning an instance of Node to this->head, the program crashes. Is there different way of assigning an instance to a pointer that was initially nullptr?
This code structure works fine on Java, I came from Java, that is why I have difficulty on C++'s pointers.
EDIT
I pasted the right code now, I'm sure. Sorry.
Ok, I have solved the problem. So, the problem is not about allocating an object to a class member, but, the problem is accessing a nullptr member: this->tail.
I edited this method, and the program now runs the way I wanted.
void addToEnd(int value) {
Node *n = new Node(value);
if(head == nullptr)
this->head = n;
else
this->tail->next = n;
this->tail = n;
}
Thanks for your help people, this question is now SOLVED. :)
I don't know about "it crashes", but the following line is not valid:
this->head = Node(value);
head is a pointer-to-Node but you're trying to assign a Node to it. Even if this automatically took the address of the temporary you created on the RHS (which it doesn't), you'd have a pointer to a local variable that doesn't exist for very long.
You should be getting a compilation error for that.
You'd have to use new to create a new object dynamically — be sure to write code to free that memory later!
You're similarly messing up dynamic memory allocation in main, where you have a needless memory leak. LinkedList list; will do fine, there.
You need to allocate memory for your Node instances. The quickest way is to call new Node(value) wherever you call Node(value). However if I were you I'd consider using shared_ptr<Node> rather than plain pointers.

Pointers - passing ptr to a ptr or passing address of a ptr

I am trying to remove the left child (10) of a sample binary search tree using two methods:
Method1: By passing pointer to a pointer to the current node.
Method2: By passing address of the pointer to the current node. This does not removes the node, but calling delete corrupts the pointer arrangement, causing a crash while printing the nodes.
The tree looks like this and I am trying to delete 10 and replace it with 5
20
|
10--|---30
|
5---|
I have some understanding of pointers. But still, I am not clear with this behavior of pointers.
#include <iostream>
class Node
{
public:
Node(int key) : leftChild(0), rightChild(0), m_key (key){}
~Node(){}
Node *leftChild;
Node *rightChild;
int m_key;
};
Node* build1234(int, int, int, int);
void print(Node *);
void print1234(Node *);
void removeLeft(Node **nodePtr)
{
Node *oldPtr = *nodePtr;
if(*nodePtr)
{
*nodePtr = (*nodePtr)->leftChild;
delete oldPtr;
}
}
int main()
{
Node *demo1 = build1234(10, 20, 30, 5);
Node *demo2 = build1234(10, 20, 30, 5);
print1234(demo1);
print1234(demo2);
//Method1 - 10 is correctly removed with 5
Node **nodePtr = &demo1;
nodePtr = &(*nodePtr)->leftChild;
removeLeft(nodePtr);
print1234(demo1);
//Method2 - 10 is not removed
Node *node = demo2;
node = node->leftChild;
removeLeft(&node);
print1234(demo2);
return 0;
}
Node* build1234(int B, int A, int C, int D)
{
Node *root = new Node(A);
root->leftChild = new Node(B);
root->rightChild = new Node(C);
root->leftChild->leftChild = new Node(D);
return root;
}
void print(Node *node)
{
if(node)
{
print(node->leftChild);
std::cout << "[" << node->m_key << "]";
print(node->rightChild);
}
}
void print1234(Node *node)
{
std::cout << std::endl;
print(node);
}
Note: This question is not about BST, but pointers. If you see the two calls to removeLeft(nodePtr) and the removeLeft(&node) in the main() function.
How are these two different?
Why the second method fails to achieve the desired result?
In the first case, you are passing an address of a pointer that exists in the tree, so you are modifying the contents of the tree directly.
In the second case, you are passing an address of a variable that is local to main() instead. The tree is not modified, and deleting from the address is accessing stack memory, which is why it crashes
You're overthinking it. All you need is a function removeLeft(Node*) that unhooks the left node and deletes it, recursively:
void removeLeft(Node * p)
{
removeBoth(p->leftChild); // recurse, OK if null
delete p->leftChild; // OK if already null
p->leftChild = 0; // necessary to make recursion terminate
}
void removeBoth(Node * p)
{
if (!p) return;
removeLeft(p);
removeRight(p);
}
If you are bad with pointers consider using smart pointers.
When using smart pointers use shared_ptr<Node> instead of Node * and make_shared(new Node); instead of new Node and remove all deletes. now you can handle pointers without caring for deletes and memory corruption.

How do you allocate memory for an linked list when passing its reference instead of its pointer?

How do you allocate memory for an link list when passing its reference instead of its pointer?
For example:
struct node {
string info;
node *next;
};
void add(node &aNode){
//if I use
node *newNode;
newNode = new node;
aNode.next = newNode; //aNode.next = newNode; doesn't work either
//allocating on heap seems to give segmentation error.
}
int main() {
node *aNode;
aNode = new node;
add (aNode);
}
Compiler error: error: invalid initialization of reference of type ‘node&’ from expr
alternatively if I use
int main() {
node aNode;
add (aNode);
add (aNode);
aNode.next->next->info = "abc";
string a = aNode.next->next->info;
}
This give segmentation fault.
So is it possible to allocate for an linked list just with its reference? (this is C++)
It should be
node * newNode = new node;
aNode.next = newNode
You have to take care of deletion manually, e.g. check if aNode.next isn't already occupied (and delete if it is).
Further, the add function signature should read:
void add(node & aNode) { ... }
By the way, the STL comes with a nice <forward_list> ;-)
It's hard to tell what you're actually asking, but going by the question title perhaps you have in mind a node structure like this:
struct Node {
Node & next;
/* payload data */
Node(Node & n) : next(n) /* ... */ { }
};
Such a node would store its successor "by reference"; but you would have to initialize it with an existing node! (There is no such thing as a "null" reference.) By the Poultry-Oval Impasse, you cannot do this.
Alright, while you continue to refuse to post your full code, here is my almost literal copy/paste of your code which works fine with me:
Update: I'm adding a feature to add a node at the end, which you might want.
#include <string>
struct node {
std::string info;
node *next;
node(std::string i = "") : info(i), next(NULL) { }
};
void add(node &aNode)
{
node *newNode;
newNode = new node;
aNode.next = newNode;
}
void add_at_end(node &aNode, std::string value = "")
{
node *newNode, *n = &aNode;
while (n->next) n = n->next; // move to the end
newNode = new node(value);
n->next = newNode;
}
int main()
{
node aNode, bNode;
add(aNode);
add_at_end(bNode, "Hello");
add_at_end(bNode, "World");
add_at_end(bNode, "!");
}
Compile with g++ -o prog prog.cpp -W -Wall -pedantic.
Finally, here's the STL way of achieving the same thing:
#include <forward_list>
#include <string>
int main() {
std::forward_list<std::string> bList;
bList.push_front("Hello");
bList.push_front("World");
bList.push_front("!");
}
In your second variant of main(), you are calling add(aNode) twice. But you're providing it the same parameter each time. So although you're creating two new node objects, one of them is lost forever (a memory leak). And aNode.next ends up pointing to the other one. aNode.next->next is not a valid pointer, hence the seg-fault when you try to access something through it.
Depending on what you want to achieve, you could try this:
node aNode;
add(aNode); // Basically does: aNode.next = new node;
add(*aNode.next); // Basically does: aNode.next->next = new node;
There are better ways of doing linked-lists, but this would at least avoid the seg-fault.
Try
int main() {
node *aNode;
aNode = new node;
add (*aNode);
}
You have to pass reference to object, not a pointer.
I checked your code and I didn't get segmentation fault when allocating on stack: http://ideone.com/gTRIG.
My proposition:
#include <string>
using namespace std;
struct node {
string info;
node *next;
node(string str): info(str), next(NULL) {}
~node() { if(next != NULL) delete next; }
node *add(string info){
node *newNode = new node(info);
return aNode.next = newNode;
}
};
int main(){
node rootNode("My rootnode");
node *nxt = rootNode.add("Next node");
nxt->add("Last node");
// No need to call delete, because destructor will clear heap
}