So I'm working on a BST, building the delete tools.
My code sequence seems to work right - Save not updating the parent or root and setting the pointer that sent it down to the deleted Node's address to NULL.
I'm passing a pointer to a pointer in my Erase and RemoveNode functions, so as to directly effect the left, right, or root data members that actually lead to the recursive call. In walking through the code, it sets *N to NULL in the remove function, but this is not reflected in the data of the calling object. Am I incorrect in using the pointer-of-a-pointer method? If so, is there a way I can recursively delete and be able to modify the prior node if the link is destroyed?
Node struct:
struct tNode
{
tNode(int n)
{
data = n;
left = NULL;
right = NULL;
}
//Works, cleans all linked objects.
//Must remember to null links when removing wanted nodes
~tNode(void)
{
//cout << "Deleting " << data << endl;
delete left;
delete right;
}
// Data members
int data;
tNode* left;
tNode* right;
};
Eraser function to recurse over the tree:
void BinSearchTree::Erase(int n, tNode** N)
{
tNode* node = *N;
if (root)
{
if (node->data > n) // post order, to avoid moving many times over
{
if (node->left)
{
Erase(n, &node->left);
}
}
else
{
if (node->right)
{
Erase(n, &node->right);
}
}
if (node->data == n)
{
RemoveNode(&node);
}
}
}
And the RemoveNode function to handle actual deletion:
void BinSearchTree::RemoveNode(tNode** N)
{
tNode* node = *N;
if (!node->left && !node->right) // is leaf
{
delete node; // remove node
size--;
*N = NULL; // null pointer for above node/structure
}
else if (!node->left) // right child
{
tNode* temp = node->right; // to strip out copied node when finished
node->data = node->right->data; // copy right node into current node
node->right = node->right->right;
node->left = node->right->left;
temp->right = NULL; // NULL because node destructor is recursive
temp->left = NULL; // ^^
delete temp;
size--;
}
else if (!node->right) // left child
{
tNode* temp = node->left; // to strip out copied node when finished
node->data = node->left->data; // copy left node into current node
node->right = node->left->right;
node->left = node->left->left;
temp->right = NULL; // NULL because node destructor is recursive
temp->left = NULL; // ^^
delete temp;
size--;
}
else // 2 children
{
tNode* temp = node->right; // find ideal child -> left-most right child
tNode* parent = NULL; // keep track of owner of ideal child
while (temp->left)
{
parent = temp;
temp = temp->left;
}
node->data = temp->data; // copy ideal child to root
if (parent)
{
parent->left = temp->right; // case that left-most child has right child of it's own
}
RemoveNode(&temp);
size--;
}
}
I think I see it. When you call RemoveNode() from Erase(), you pass it the value &node. Pass in N instead.
Here's what's happening: The line tNode* node = *N; creates a local variable on the stack. A copy of *N. And while that variable has the same value as *N (at first), it's stored in a different place in memory: node is on the stack, and *N is somewhere in the heap.
So, since you pass &node to RemoveNode(), node is what gets changed. Not *N - it's somewhere else. But if you pass in, you're changing what you want to.
Hope that helps!
PS: If that's not clear, let me know! Writing about double-pointers might just be harder than using them...
Save yourself the trouble of using double pointers and just use a reference to pointer. They have the same semantic of normal pointers, but assigning them will actually change the passed pointer.
void BinSearchTree::Erase(int item, tNode*& node);
Also, use expressive names and save the single letter variable names for loops.
Related
In place of *head_ref = temp->next;, why can't I assign it as *head_ref = *head_ref->next?
Why should I use temp? Aren't they pointing to the same place?
class Node{
public:
int data;
Node* next;
};
void deleteNode(Node** head_ref, int key){
Node* temp = *head_ref;
Node* prev = NULL;
if(temp!=NULL && temp->data==key){
*head_ref = temp->next;
delete temp;
return;
}
else{
while(temp!=NULL && *head_ref->data!=key){
prev = temp;
temp = temp->next;
}
}
Your code does not compile, *head_ref->data should be (*head_ref)->data.
The reason why you should use temp is that you want to modify *head_ref only if the element you want to delete is the head element. If you delete any other element of the list, the head pointer must stay the same.
But your code is wrong anyway. You're doing things in the wrong order. You must first find the element you want to delete, and then handle the deletion.
Your code handles the deletion first and then finds the element to delete which is absurd.
You want this:
void deleteNode(Node** head_ref, int key) {
Node* current = *head_ref;
Node* previous = NULL;
// find element to delete
while (current && current->data != key)
{
previous = current;
current = current->next;
}
// if current is NULL here then the element has not been found
if (current != NULL)
{
// element found,
// current points to element found
// previous points to previous element or NULL if current is head
if (previous == NULL)
{
// deleting head element -> we need to update head_ref
*head_ref = current->next;
}
else
{
// deleting any other element -> update next pointer of previous element
previous->next = current->next;
}
delete current;
}
}
That being said, this is rather C code than C++ code. You should use standard containers rather than making your own, or at least use C++ idioms such as constructors.
I have a function for finding a specified node in a binary search tree as well as that nodes previous node:
bool collection::removeFromTree(const char name[])
{
treeNode ** prev = nullptr;
for (treeNode ** curr = &root; *curr;)
{
int8_t result = strcmp(name, (*curr)->item->getName());
if (result == 0)
{
deleteNode(prev, curr);
return true;
}
else if (result < 0)
{
prev = curr;
curr = &((*curr)->left);
}
else if (result > 0)
{
prev = curr;
curr = &((*curr)->right);
}
}
return false;
}
The problem I'm having is not with this function, but with my deleteNode() function. I'm unable to assign my previous node to point to the node after my current node (or the node I plan to delete). This is the important part of my deleteNode() function:
void collection::deleteNode(treeNode **& prevNode, treeNode **& goneNode)
{
//irrelevant code
//if it has right child
if (!(*goneNode)->left)
{
treeNode * temp = (*goneNode)->right;
(*goneNode)->right = nullptr;
delete *goneNode;
(*prevNode)->right = temp;
*goneNode = nullptr;
}
//irrelevant code
}
The problem, of course, is that (*prevNode)->right becomes null after this function runs. The pointer to the pointer goes out of scope and the data is lost. Any nodes that were to the right of goneNode go out of scope. Is there any nice way to fix this?
I've also tried:
void collection::deleteNode(treeNode **& prevNode, treeNode **& goneNode)
{
//irrelevant code
//if it has right child
if (!(*goneNode)->left)
{
(*prevNode)->right = (*goneNode)->right;
(*goneNode)->right = nullptr;
delete *goneNode;
*goneNode = nullptr;
}
//irrelevant code
}
When I do that, (*prevNode)->right becomes null right after I delete *goneNode; (before the function even finishes executing).
The advantage to using a two-star pointer here was to avoid using a current and previous pointer. Having a pointer to the pointer to a node, we can edit the link that bound that node into the tree. The rest of the previous node doesn't matter; we only care about the pointer to what I'll call the bad_node (the one to remove).
(*prevNode)->right = (*goneNode)->right; //50% chance that (*prevNode)->right was (*goneNode)
Either bad_node was to the right of prevNode, and we erase the location of bad_node, or it was to the left and we mutilate the tree. I'm betting in your scenario that the first one is happening. Once that line executes, *goneNode is going to be the wrong value, and so it ends up leaking goneNode and deleting its child.
To rip myself off a little:
else if (!(*goneNode)->left) //one child on right
{
treeNode* temp = *goneNode; //store address of bad_node
*goneNode = (*goneNode)->right; //replace bad_node with child
temp->right = nullptr; //bad_node now has no children
delete temp; //delete bad_node
}
The important thing to remember is that *goneNode is the pointer to bad_node in the tree. Assigning to *goneNode changes the structure of the tree.
I have these functions to remove a node from my binary search tree:
bool collection::removeFromTree(const char name[])
{
for (treeNode * curr = root; curr;)
{
int8_t result = strcmp(name, curr->item->getName());
if (result == 0)
{
deleteNode(curr);
return true;
}
else if (result < 0)
curr = curr->left;
else if (result > 0)
curr = curr->right;
}
return false;
}
void collection::deleteNode(treeNode *& goneNode)
{
//if it's a leaf
if (!goneNode->left && !goneNode->right)
{
delete goneNode; //node's destructor gets invoked
goneNode = nullptr;
}
//if it has right child
else if (!goneNode->left)
{
goneNode = goneNode->right;
}
//if it has left child
else if (!goneNode->right)
{
goneNode = goneNode->left;
}
//if it has both children
else
{
treeNode * prev = nullptr;
treeNode * curr = goneNode->right;
while (curr->left)
{
prev = curr;
curr = curr->left;
}
//prev points to the copy over data
delete goneNode->item;
if (!prev)
{
goneNode->item = curr->item;
goneNode->right = curr->right;
curr->item = nullptr;
}
else
{
goneNode->item = curr->item;
curr->item = nullptr;
prev->left = curr->right;
}
}
}
This runs fine, but when I try to list all the elements in my tree after deleting a node (with these functions):
void collection::displayByName() const
{
std::cout << std::endl
<< "========================================" << std::endl;
//display the tree inorder
listAll(root);
}
void collection::listAll(const treeNode * const & root) const
{
if (root)
{
std::cout << *(root->item) << std::endl
<< "========================================" << std::endl;
listAll(root->left);
listAll(root->right);
}
}
I receive this error:
And when I quit the program after deleting a node (invoking these destructors):
collection::~collection()
{
delete root;
}
collection::treeNode::~treeNode()
{
delete left;
delete right;
}
I recieve this error:
Any suggestions would be greatly appreciated because I see no reason for my listAll() function to be calling nodes that I've already deleted.
By the way, this is my struct for my treeNode:
struct treeNode
{
treeNode();
treeNode(vendor *& item);
~treeNode();
vendor * item;
treeNode *left, *right;
};
treeNode * root; //the bst
hashNode ** table; //the hash table
uint8_t capacity;
uint8_t size;
const static uint8_t INIT_CAP = 20;
When you need to remove a node from a singly linked list or a tree, I find using a pointer to pointer is handy. Namely, if we have a treeNode** ptr;, then *ptr is the pointer to our node. So, if ptr = &root, then *ptr = nullptr sets root to nullptr.
I removed the deleteNode function and threw its logic in the removeFromTree function.
bool collection::removeFromTree(const char name[])
{
treeNode** ptr = &root;
Instead of being a pointer to treeNode, ptr will point to a treeNode* inside the tree structure. This way, we can modify the pointer that led us to the current node. The lines marked //same as before have the same logic you were using, just possibly modified to account for the fact ptr has another level of dereferencing to do.
int result; //same as before
while (*ptr) //While we haven't hit a dead end
{
result = strcmp(name, (*ptr)->item->getName()); //same as before
if (result < 0) //same as before
ptr = &((*ptr)->left); //same as before
else if (result > 0) //same as before
ptr = &((*ptr)->right); //same as before
else //begin deleteNode() logic
{
if ((*ptr)->left && (*ptr)->right) //two children
{
Here, we use pointers to member because the alternative was a conditional operator on every line. If a node has two children, we need to find either the rightmost node on the left side, or the leftmost node on the right side. That's the node we can replace the current node with.
treeNode* treeNode::*dir = some_condition ? &treeNode::right : &treeNode::left; //pointer to treeNode member of type treeNode*
treeNode* treeNode::*ndir = some_condition ? &treeNode::left : &treeNode::right; //pointer to treeNode member of type treeNode*
dir now either points to left or right, which is the direction we are searching the tree for. ndir is the opposite direction. So, if we want the rightmost node on the left side, (*ptr)->*dir == (*ptr)->left and (*ptr->*ndir == (*ptr)->right. If we want the leftmost right node, it would be reversed. This is just a more complicated way to do less work, really. It shouldn't be hard to remove. some_condition is just either true or false. true means the left side of the tree (from the current node) loses a node, and false means the right side does.
treeNode** replacement = &((*ptr)->*ndir); //the node to replace the current one with
while ((*replacement)->*dir) //While we aren't at the edge
replacement = &((*replacement)->*dir);
This loops until *replacement is the node we need to replace *ptr with.
treeNode* rep_branch = (*replacement)->*ndir; //If the replacement node had a child, this is now it
(*replacement)->left = (*ptr)->left; //Copy current to replacement
(*replacement)->right = (*ptr)->right; //Copy current to replacement
(*ptr)->left = nullptr; //null out current in case of destructor
(*ptr)->right = nullptr; //null out current in case of destructor
Now, the replacement node is pointing to the node-to-be-deleted's children, and our soon to be expired node has no children anymore. Now, it's safe to delete the unwanted node. If the node class had a destructor to delete its children, the left and right pointers were set to nullptr just in case.
delete *ptr; //delete unwanted node
*ptr = *replacement; //replacement node has taken the unwanted node's place in the tree
*replacement = rep_branch; //The replacement's child takes its own place
}
This completes the tree's structure. Wherever the unwanted node was, the replacement node has taken its place. And because the replacement node was required to be an edge node, it had at most one child. We just replace it with the child.
else if ((*ptr)->left) //one child on left
{
treeNode* current = *ptr;
*ptr = (*ptr)->left; //replace current with left
current->left = nullptr; //null out for safety
delete current;
}
else if ((*ptr)->right) //one child on right
{
treeNode* current = *ptr;
*ptr = (*ptr)->right; //replace current with right
current->right = nullptr; //null out for safety
delete current;
}
else //no children
{
delete *ptr;
*ptr = nullptr;
}
return true; //yay it's over
}
}
return false; //never found it
}
The rest is fairly straightforward, just replacing easier nodes and returning. Hopefully this gives you some ideas about how to approach problems like this, and the occasional uses of some of these structures. This is what I meant about using treeNode** over treeNode* for operations like this.
I've checked the boards and could not find any help with this. I find it easy to implement recursive functions given base and general cases, but this doesn't work the way I do it. I'm supposed to iterate down a list until I reach the tail of a linked list. If the next node is NULL, then I have to store the value at the last node, remove that node, and return the value. So it's similar to a dequeue method, except it's performed recursively. What am I doing wrong?
int LinkedList::removeTailRec(Node *n)
{
// check for the base case(s)
if(n->next == NULL)
{
Node *tmp = new Node();
tmp = n;
int val = n->value;
tmp = NULL;
return val;
}
else
return removeTailRec(n->next);
// else call the recursive method
}
First, I recommend you use nullptr instead of NULL.
Then, onto your code. You're actually not removing anything from your list.
if(n->next == NULL)
{
Node *tmp = new Node();
^^^^^^^^^^
//Useless, and dangerous. This memory is never free'd
tmp = n;
int val = n->value;
tmp = NULL;
^^^^^^^^^^
//You just set a local variable to NULL, you're not deleting anything
return val;
}
If you want to remove the node, you'll have to keep a reference to the previous node (e.g. having a doubly linked list, that is, having a pointer to the next element and a pointer to the previous element in each node, or working on the previous node directly).
Set this previous node's next to nullptr, store the node's value and then delete the Node pointer.
One way to do this is to work with the pointer to the next node :
int LinkedList::removeTailRec(Node *n)
{
//EDIT: Adding a check for n validity
if(!n){
//Here, you should have a way of detecting
//a call to your method with a null pointer
return 0;
}
Node* nextNode = n->next;
// check for the base case(s)
if(nextNode->next == nullptr)
{
//Get the next node value
int val = nextNode->value;
//Set the current node next member to nullptr
n->next = nullptr;
//Free the last node
delete nextNode;
return val;
}
else{
return removeTailRec(n->next);
}
// else call the recursive method
}
You are storing the result but not deleting it from linked list. You can return result in another variable (pointer : result).
Node* getTail(Node *n,int *result){
//u can even free the memory
if(!n->next)
{
result=n->value;
return NULL;
}
n->next=getTail(n->next,result);
}
or you can do it other way
int getTail(Node *n)
{
if(!n) return 0;
if(n->next)
{
if(!n->next->next)
{
Node *frnode=n->next;
int result=n->next->value;
n->next=NULL;
delete frnode;
return result;
}
getTail(n->next);
}
You are not removing last node in your code, and you leak another (temporary) node here.
To remove last node you have to zero the link in the previous node.
Your code should look like
...
if (n == NULL || n->next == NULL)
throw std::out_of_range("node");
if(n->next->next == NULL)
{
int val = n->next->value;
delete n->next;
n->next = NULL;
return val;
}
else ...
Be aware of the fact that c++ is not a functional language and has no optimizations for tail recursion, so in real application as your lists grow big enough you'll eventually have failure with stack overflow =) use Haskell or Erlang for this style of programming, in c++ use for or while.
You should set the Node n's previous Node's next field to NULL when n is the tail Node.
I am trying to create my own datatype that is like a vector or an array.
I am having troubles with my print function; When I go to print the list, it only prints the last item in the list.
// LinkedListClass.cpp : Defines the entry point for the console application.
#include "stdafx.h"
#include <iostream>
class Node
{
public:
int value;
Node* next;
Node::Node(int val)
{
value = val;
};
};
class List
{
public:
Node* firstNode;
Node* currentNode;
int size;
List::List()
{
firstNode = NULL;
currentNode = firstNode;
size = 0;
};
void push(Node* node)
{
if(firstNode == NULL)
{
firstNode = node;
firstNode->next = currentNode;
size++;
}
else
{
currentNode = node;
currentNode = currentNode->next;
size++;
}
};
void print()
{
if(firstNode != NULL)
{
Node* printNode = firstNode;
while(printNode->next != NULL)
{
std::cout << "List Item " << printNode->value << std::endl;
printNode = printNode->next;
}
}
};
};
int _tmain(int argc, _TCHAR* argv[])
{
List ll = List();
for(int i = 0; i < 10; ++i)
{
Node val = Node(i);
ll.push(&val);
}
std::cout << ll.firstNode->value << std::endl;
ll.print();
std::cout << "Size " << ll.size << std::endl;
std::cin.ignore();
return 0;
}
/* Output
9
Size 10
*/
I know this is nowhere near completed, but if you have any other pointers (lol), please feel free to suggest.
There are three important errors:
push() --- fixed
void push(Node* node)
{
if(firstNode == NULL)
{
firstNode = node;
currentNode = node;
// firstNode->next = currentNode; --> this does nothing useful!
size++;
}
else
{
currentNode->next = node;
currentNode = node;
//currentNode = node; -|
//currentNode = currentNode->next; -|----> why? what? Do explain.
size++;
}
}
I think by assigning firstNode->next = currentNode; you expected the next time currentNode was updated, it would update firstNode->next as well.
It doesn't work that way.
firstNode->next = currentNode; implies that the address stored in currentNode is now in firstNode->next. So next time you store something in currentNode = node; you're not storing it in firstNode->next. So you have a broken linked list --- which is why your output didn't go very far.
Also, this is really bad. By setting currentNode=node before setting the current node's next pointer to node, you've broken the list again. You should first point currentNode->next to node and then set the currentNode as node (node being the node which you're pushing onto your list).
Node val = Node(i);
The scope of val is only within that iteration of your loop. Once you loop around, it's off the stack and doesn't exist anymore. But you've copied the pointer of val to your list --- so now with the right push method, you're just adding a dangling pointer.
Node *val = new Node(i);
ll.push(val);
You need to put it on the heap so it stays on till you don't need it anymore.
... which leads us to your destructor!
Since you've allocated a node, you'll need to deallocate it. So do that in your destructor --- traverse your list and deallocate all those nodes.
The following lead to undefined behavior:
Node val = Node(i);
ll.push(&val); // take address of temporary
...
firstNode = node; // store address of temporary here
...
ll.print(); // temporary `val` was destroyed, but all nodes are point to it
You could change your code as follows:
Node* val = new Node(i);
ll.push( val );
And don't forget to delete all nodes later.
Your push() method is incorrect. The first time you push a node, it correctly assigns it to firstNode, but every subsequent push() just sets currentNode to the new node, and then sets currentNode to NULL -- you're not actually adding anything to your list.
I think it bears mentioning that pointers are not reference-by-name in C++. For instance, setting firstNode->next = currentNode doesn't make currentNode the next element in the list; it just makes firstNode->next point to the same address that currentNode does (in this case, NULL).
I'm not going to write the code for you, but here's how your push() function should work. The key is that you should be setting the 'next' field of an existing node to your new node, rather than currentNode to the new node:
In the case where firstNode is NULL,
set firstNode to the new node and
set firstNode->next to NULL (since
it has no next element). You can
also set currentNode = firstNode
here for convenience.
In the case where firstNode is not
NULL, we need to walk from firstNode
down the chain until we find a node
whose next field is NULL, then set
its next field to the new node.
Alternatively, we can use that
currentNode pointer to access the
last element in list and do the same
thing, being sure to set currentNode
to point to the new node when we're
done.
You basically have part 1 done, but part 2 still needs to be implemented. Feel free to ask for clarification/give criticism. :)
try it like Node* val=new Node(i)
previously u were storing the temporary variable. so no store the ndoe in dynamic memory so seprate memory can be given.
when u were creating the node it is create for temparary purpose
&temporary address were stored so when u traverse back the temporary memory had been released u will find there some garbage. value.