remove item on linked list - c++

Question

I have implemented a linked list but I find that removal of the last element in the linked list always fails. I checked my code several times but didn't find the logic mistake. Remove from the head and middle node works fine. Below is my code:
#include<iostream>
using namespace std;
template<typename T>
class Node{ //
private:
T val;
Node *next;
Node *prev;
public:
Node(T value){
val = value;
next = nullptr;
prev = nullptr;
}
~Node(){
if(prev != nullptr){
prev->next = next;
}
if(next != nullptr){
next->prev= prev;
}
}
T& getVal(){ //
return val;
}
Node* getNext(){
return next;
}
Node* getPrev(){
return prev;
}
//insert node after this node
void insert(Node *n){
if(next != nullptr){
next->prev = n;
}
n->next = next;
n->prev = this;
next = n;
}
void deleteNode(){
if(prev != nullptr && next != nullptr){
prev->next = next;
next->prev = prev;
}
}
};
//build a linked list with deletion and push back function
template<typename T>
class LList{
private:
Node<T> *head;
Node<T> *tail;
public:
LList(){
head = nullptr;
tail = nullptr;
}
~LList(){
if(nullptr != head){
while(head->getNext() != nullptr){ //
delete head->getNext();
}
delete head;
}
}
void push_back(T val){
Node<T>* n = new Node<T>(val);
if(head == nullptr){
head = n ;
tail = head;
}
else{
tail->insert(n);
tail = n;
}
}
void deleteItem(T item){
Node<T> *node = head;
//delete head
if(node->getVal() == item){
head = node->getNext();
node->deleteNode();
return;
}
//delete middle and tail
while(node->getVal() != item && node->getNext() != nullptr ){
node = node->getNext();
}
if(node->getVal() == item && node == tail){
tail = node->getPrev();
node->deleteNode();
return;
}
if(node->getVal() == item && node != tail){
node->deleteNode();
return;
}
else {
cout<<"didnt find the item "<<item<<endl;
}
}
void print(){
Node<T> *node = head;
while(node->getNext() != nullptr){
cout<<node->getVal()<<endl;
node = node->getNext();
}
cout<<node->getVal()<<endl;
}
};
int main(){
LList<double> list;
list.push_back(3.13);
list.push_back(2.8);
list.push_back(23);
list.push_back(4);
list.print();
list.deleteItem(3.13);
list.deleteItem(2);
list.deleteItem(4);
list.print();
return 0;
}

Answer 1

Here it is:
void deleteNode(){
if(prev != nullptr && next != nullptr){
prev->next = next;
next->prev = prev;
}
}
};
For the last element i guess that next equals nullptr and so the whole condition fails.
What I don't understand, is why removing the first element doesn't fail too, you probably should x-check your code for this.
[edit]
Here's my solution to fix this bug, it's basically the same as in the destructor for node:
void deleteNode(){
if(prev != nullptr){
prev->next = next;
}
if(next != nullptr){
next->prev= prev;
}
};
I think all cases have been considered in this:
Head := prev == null & next->prev = nullptr (assuming prev for this node was nullptr before)
Middle := prev->next = next & next->prev = prev
Tail := prev->next = null (assuming as above) & next == null
Head&Tail := prev == null & next == null (Only element of the list is deleted, so no references to be changed)
[/edit]