I wanna ask what the time complexity of post traversal in BST if I removed some specific nodes in it while doing the traversal, for example: A dictionary BST that has a word, and a meaning. We want to delete all words with data equal to an empty string, the tree could be unbalanced.
this is my algorithim:
void delete_nodes(Node*& node) {
if (node == nullptr) {
return;
}
delete_nodes(node->left);
delete_nodes(node->right);
if (node->data == "") {
// Node to be deleted has 0 or 1 child
if (node->left == nullptr) {
Node* temp = node->right;
delete node;
node = temp;
} else if (node->right == nullptr) {
Node* temp = node->left;
delete node;
node = temp;
} else {
// Node to be deleted has 2 children
// Swap the values of the node and the leftmost node in its right subtree
Node* temp = node->right;
while (temp->left != nullptr) {
temp = temp->left;
}
node->data = temp->data;
// Set the data of the leftmost node in the right subtree to an empty string
temp->data = "";
// Delete the leftmost node in the right subtree
delete_nodes(node->right);
}
}
}
Related
void deletenode(string key) {
if (last == NULL) {
cout << "your circular linked list is an empty one" << endl;
}
else {
node* p = last->next;
node* prev = last;
do {
if (p->title == key) {
node* temp = p;
prev->next = p->next;
delete(temp);
}
else {
p = p->next;
prev = prev->next;
}
} while (p != last->next);
}}
I was trying to delete a node with key value. For instance, if node p->title is my key then I want to delete that node. However, I implemented it with other values but the code doesn't seem to work or delete a node with key value from my circular linked list. What is the mistake in the function?
circular linked list value "cat", "dog", "rat", "horse", the key to be deleted was "dog". When I traverse throughout the linked list after the deletion it still printed everything including "dog", which means deletion didn't work.
Anytime you write a "delete from the linked list" function, you have to account for the possibility that you are deleting from the "head" or whatever pointer you are referencing with the list. The common pattern is for the function to return the new head of the list if it changed, else return the current head.
Node* deletenode(Node* head, const string& key) {
// empty list
if (head == nullptr) {
return nullptr;
}
// single item list
if (head->next == nullptr || head->next == head) {
if (head->title == key) {
delete head;
head = nullptr;
}
return head;
}
// two or more item list, possibly circular
Node* prev = head;
Node* current = head->next;
Node* first = current;
while (current && current->title != key) {
prev = current;
current = current->next;
if (current == first) {
break;
}
}
if (current == nullptr || current->title != key) {
return head; // not found
}
prev->next = current->next;
if (current == head) {
head = current->next;
}
delete current;
return head;
}
I don't see the full code so I can't make a comment I tried to implement the function, hope it helps you with the comments.
void deleteNodeWithKey(node* head, string key)
{
node *curr = head;
node *last , *temp;
//Search for last node
while (curr->next != head)
{
curr = curr->next;
}
last = curr;
//If head is the desired key, make head's next new head
//and connect last node to new head
if (head->key == key)
{
temp = head->next;
delete head;
head = temp;
last->next = head;
return;
}
temp = head->next;
//Search for node with the given key
node *prev = head;
while (temp != head)
{
if (temp->key == key)
{
prev->next = temp->next;
delete temp;
return;
}
temp = temp->next;
prev = prev->next;
}
//If function gets here, key was not found
}
I made some changes to your code
void deletenode(string key) {
if (last == NULL) {
cout << "your circular linked list is an empty one" << endl;
}
else {
node* prev = last;
// If head is to be deleted
if (last->title == key) {
while (prev->next != last)
prev = (prev)->next;
prev->next = last->next;
free(last);
last = prev->next;
return;
}
node* p = last->next;
do {
if (p->next->title == key) {
node* temp = p->next;
p->next = temp->next;
delete(temp);
}
else {
p = p->next;
prev = prev->next;
}
} while (p != last->next);
}
}
Having a problem with linked list. Need to create a method, which will replace data in list, by not creating a new element but by changing pointers. For now I have such method:
void replaceValues(Node* head, int indexOne, int indexTwo)
{
Node* temporaryOne = NULL;
Node* temporaryTwo = NULL;
Node* temp = NULL;
Node* current = head;
int count = 0;
while (current != NULL) {
if (count == indexOne)
{
temporaryOne = current;
}
else if (count == indexTwo)
{
temporaryTwo = current;
}
count++;
current = current->next;
}
current = head;
count = 0;
while (current != NULL) {
if (count == indexOne)
{
head = temporaryTwo;
}
else if (count == indexTwo)
{
head = temporaryOne;
}
count++;
current = current->next;
}
}
I am sure, that exists a more simpler way, how to do it, but I don't fully understand, how it works...
Thanks in advance for help.
I assume that with "replace" you actually mean "swap"/"exchange".
Some issues:
The argument head should be passed by reference, as one of the nodes to swap may actually be that head node, and then head should refer to the other node after the function has done its job.
The node before temporaryOne will need its next pointer to change, so you should stop your loops one step earlier in order to have access to that node and do that.
In some cases head may need to change, but this is certainly not always the case, so doing head = temporaryOne or head = temporaryTwo is certainly not right. In most cases you'll need to link to the swapped node from the preceding node (see previous point).
The next pointer of the node that is swapped will also need to change, as the node that follows it will be a different one than before.
As mentioned already in comments, it is advised to split the task into removals and insertions, as the fiddling with next pointers can get confusing when you try to cover all possible cases, notably making the distinction between the case where the two nodes are adjacent and when they are not.
Here are some functions that split the work into removal, insertion and finally exchanging nodes:
Node* removeNode(Node* &head, int index) {
// If index is out of range, no node is removed, and function returns nullptr
// Otherwise the extracted node is returned.
if (head == nullptr || index < 0) return nullptr;
Node* current = head;
if (index == 0) {
head = head->next;
current->next = nullptr;
return current;
}
while (--index > 0) {
current = current->next;
if (current == nullptr) return nullptr;
}
Node* temp = current->next;
if (temp != nullptr) {
current->next = temp->next;
temp->next = nullptr;
}
return temp;
}
void insertNode(Node* &head, Node* node, int index) {
// If index is too large, node is inserted at the end of the list
// If index is negative, node is inserted at the head of the list
if (index <= 0 || head == nullptr) {
node->next = head;
head = node;
return;
}
Node* current = head;
while (--index > 0 && current->next != nullptr) {
current = current->next;
}
node->next = current->next;
current->next = node;
}
bool exchangeNodes(Node* &head, int indexOne, int indexTwo)
{
// Returns true when successful, false when at least one index
// was out of range, or the two indexes were the same
if (head == NULL || head->next == NULL || indexOne == indexTwo || indexOne < 0) return false;
// To ensure the right order of operations, require the first index is the lesser:
if (indexOne > indexTwo) return exchangeNodes(head, indexTwo, indexOne);
Node* two = removeNode(head, indexTwo);
if (two == nullptr) return false; // out of range
Node* one = removeNode(head, indexOne);
insertNode(head, two, indexOne);
insertNode(head, one, indexTwo);
return true;
}
This is a basic function that takes an iterator position and deletes the node in this position but it gives me a runtime error. what am i doing wrong?
iterate erase(iterate position)
{
iterate i;
Node<T>* temp = head;
if (head == NULL) {
cout << "empty list" << endl;
}
else if (position.pointer == head) {
head = temp->next;
temp->next->previous = NULL;
delete position.pointer;
}
else {
while (temp != NULL) {
if (temp == position.pointer->previous) {
temp->next = position.pointer->next;
temp->next->previous = temp;
i.pointer = temp->next;
delete position.pointer;
return i;
}
}
}
Your function is lacking adequate return statements. There are multiple flows that can cause the function to exit, but only one of them has a return statement. So the return value will largely be indeterminate, causing undefined behavior for any caller that tries to use the return value.
In any case, your while loop iterates forever, because you are not updating temp on each iteration of the loop. You also have a NULL pointer dereference if position is pointing at the last node in the list, as you are not checking the new temp->next for NULL before accessing temp->next->previous.
But, you really don't need the while loop at all. The thing about a double-linked list is that, given any node in the list, you have direct access to the nodes that are surrounding it on both sides. So there is no need to iterate the list hunting for nodes.
Try something more like this instead:
iterate erase(iterate position)
{
Node<T> *temp = position.pointer;
if (!temp) return end();
Node<T> *next = temp->next;
Node<T> *previous = temp->previous;
if (next) next->previous = previous;
if (previous) previous->next = next;
if (temp == head) head = next;
//if (temp == tail) tail = previous;
delete temp;
iterate i;
i.pointer = next;
return i;
}
Alternatively:
iterate erase(iterate position)
{
Node<T> *temp = position.pointer;
if (!temp) return end();
Node<T> *dummy; // <-- only if no tail ...
Node<T> **previous = (temp->next) ? &(temp->next->previous) : &dummy/*&tail*/;
Node<T> **next = (temp->previous) ? &(temp->previous->next) : &head;
*previous = temp->previous;
*next = temp->next;
delete temp;
iterate i;
i.pointer = *next;
return i;
}
I'm having difficulty deleting a node in my binary search tree. The delete function is part of my Node class, and my findMin function is as well. Below is my delete function...
/**********************************************
* Delete
**********************************************/
node* node::Delete(node *root, string stuff)
{
//node *temp;
if (root == NULL) // Searches for value in tree
return NULL;
if (stuff < root->val) // String is in left subtree
root->left = Delete(root->left, stuff);
else if (stuff > root->val) // String is in right subtree
root->right = Delete(root->right, stuff);
else
{ // No children
if ((root->left == NULL) && (root->right == NULL))
{
delete(root);
root = NULL;
}
else if ((root->right == NULL) && (root->left != NULL)) // One left child node
{
node *temp = root;
root = root->left;
delete temp;
temp = NULL;
}
else if ((root->left == NULL) && (root->right!= NULL)) // One right child node
{
node *temp = root;
root = root->right;
delete temp;
temp = NULL;
}
else // Two children
{
node *temp = findMin(root->right); // Finds smallest value in right subtree
root->val = temp->val;
root->right = Delete(root->right, temp->val);
}
}
return root;
}
Below is my Destructor, which is giving me a SIGABRT (I'm using Xcode)
/**********************************************
* Destructor
**********************************************/
node::~node()
{
if (left != NULL) delete left;
if (right != NULL) delete right;
}
What my code is actually doing is not only deleting the node I intend to delete, but its child node. What could I be doing wrong? Is it an error with memory allocation? Is it an error with how I set the value to the child node?
You need to null your pointers to left and right before deleting a node.
You call:
node *temp = root;
root = root->left;
delete temp;
temp = NULL;
When you "delete temp" you are deleting a node which still points to root->left and root->right and your destructor insures they are also removed. You should instead do something like this:
node *temp = root;
root = root->left;
temp->left = NULL;
temp->right = NULL;
delete temp;
temp = NULL;
Also in your destructor you don't need to check if they are equal to null since delete already preforms this check.
I have a recRemove function that recursively removes the given node. I also have a findMin function that finds the smallest node in the BST. I'm having trouble merging the two so that I can remove the smallest(or largest) node. This is what I tried doing but it just returned garbage: Full code: https://pastebin.com/HCVsUZ4S
//remove min node in BST
node * extractMin()
{
return recRemove(root, findMin(root));
}
//delete node from tree
node * recRemove(node * root, double data)
{
//3 cases: no children, one child, 2 children
if (root == NULL)
{
return NULL;
}
else if (data < root->data)
{
root->left = recRemove(root->left, data);
}
else if(data > root->data)
{
root->right = recRemove(root->right, data);
}
else
{
if (root->right == NULL && root->left == NULL) //no children
{
delete root;
root = NULL;
return root;
}
else if(root->left == NULL) //only right child
{
temp = root;
root = root->right;
delete temp;
return root;
}
else if(root->right == NULL) //only left child
{
temp = root;
root = root->left;
delete temp;
return root;
}
else //2 children
{
temp->data = findMin(root->right);
root->data = temp->data;
root->right = recRemove(root->right, temp->data);
}
}
return root;
}
//find min node in BST
double findMin(node * p)
{
if(p == NULL)
{
return -1;
}
else
{
//in a balanced BST the minimum node is the leftmost node so,
//we traverse the left subtree until we get to the leftmost node and return and remove it.
temp = p;
while(temp->left != NULL)
{
temp = temp->left;
}
return temp->data;
}
}
sorry , can't write comments yet (will delete this later)
Where is temp defined? If it is a global variable than this is probably the issue...
Edit:
Have now seen the pasebin.....
temp is a member variable. Change it to a local variable.
Make sure to delete it before leaving the function. (best use std::unique_ptr<>)