How does the recursion work for the function provided?
struct tree
{
int value;
struct tree* left;
struct tree* right;
};
void recursion(tree* head)
{
if (head != NULL)
{
recursion(head->left);
cout << head->value << endl;
recursion(head->right);
}
}
int main()
{
tree* head;
tree* newNode = new(struct tree);
newNode->value = 2;
newNode->left = NULL;
newNode->right = NULL;
head = newNode;
newNode = new(struct tree);
newNode->value = 1;
newNode->left = NULL;
newNode->right = NULL;
head->left = newNode;
newNode = new(struct tree);
newNode->value = 3;
newNode->left = NULL;
newNode->right = NULL;
head->right = newNode;
recursion(head);
return 0;
}
Traversing through the tree node with the recursion() function, during recursion(head->left) inside the function, when it finally hits NULL does the end of that state of recursion become 'head->value' and use the cout << head->value << endl; line of code to print 1? Vice versa for recursion(head->right)? I'm having a hard time understanding how exactly the last left node is being printed granted i'm only using cout for the head->value node.
Using a debugger makes it easier to learn how code is executed as it follows the exact execution order. This when threading is not taken into account.
Sometimes, with difficult patterns, it helps to just draw it on paper. It will make is very visible for what is going on.
The graph below starts with your first created struct with newNode->value = 2;
The green arrows show the order of execution.
out 1, out 2 and out 3 are the head->value in your function.
I wrote an AVL tree, I'm sure that it works correctly, but my vertex counting does not work for the subtree. Help find a mistake.
I am sure that the error is in this part of the delete function, if this is not enough, I will write its full part.
If I’m fundamentally wrong, then tell me how to implement the algorithm for finding the number of vertices in a subtree for an AVL tree
class Node
{
public:
int key;
Node *left;
Node *right;
int height;
int kol; // number of vertices in the subtree
};
Node* newNode(int key){
Node* node = new Node();
node->key = key;
node->left = nullptr;
node->right = nullptr;
node->height = 1;
node->kol = 1;
return node;
}
Node* minValueNode(Node* node){
Node* current = node;
while (current->left != nullptr)
current = current->left;
return current;
}
Node* deleteNode(Node* root, int key)
{
if (root == NULL)
return root;
if ( key < root->key ){
root->left = deleteNode(root->left, key);
root->kol--;
}
else if( key > root->key ){
root->right = deleteNode(root->right, key);
root->kol--;
}
else
{
if( (root->left == NULL) ||
(root->right == NULL) )
{
Node *temp = root->left ?
root->left :
root->right;
if (temp == NULL)
{
temp = root;
root = NULL;
}
else
{
int t = root->kol;
*root = *temp;
root->kol = t - 1;
}
free(temp);
}
else
{
Node* temp = minValueNode(root->right);
root->key = temp->key;
//root->kol = temp->kol;
root->right = deleteNode(root->right,
temp->key);
}
}
if (root == NULL)
return root;
return root;
}
Here using the function returnReverseLinkedList I am returning the reversed linked list of the given linked list. But the problem with this approach is that i lose the original linked list. So I make another fucntion called createReversedLinkedList to make a copy of the original linked list and reverse the copy and maintain possession of both.
unfortunately createReversedLinkedList is giving Runtime error.
obviously my end goal is to check if the given linked list is palindrome or not. This issue is just a stepping stone.
Could someone tell me why?
//Check if a linked list is a palindrome
#include <iostream>
using namespace std;
class node
{
public:
int data;
node *next;
node(int data)
{
this->data = data;
this->next = NULL;
}
};
node *returnReverseLinkedList(node *head)
{
// Will Lose original Linked List
if (head == NULL)
return NULL;
else if (head != NULL && head->next == NULL)
return head;
node *prev = NULL;
node *curr = head;
node *tempNext = head->next;
while (tempNext != NULL)
{
curr->next = prev;
prev = curr;
curr = tempNext;
tempNext = tempNext->next;
}
curr->next = prev;
return curr;
}
node *createReversedLinkedList(node *head)
{
if (head == NULL)
return NULL;
else if (head != NULL && head->next == NULL)
return NULL;
else
{
node *temp = head;
node *newHead = NULL;
node *newTail = NULL;
while (temp != NULL)
{
node *newNode = new node(temp->data);
if (newHead == NULL)
{
newHead = newNode;
newTail = newNode;
}
else
{
newTail->next = newNode;
newTail = newNode;
}
}
return returnReverseLinkedList(newHead);
}
}
bool check_palindrome(node *head)
{
node *original = head;
node *reverse = returnReverseLinkedList(head);
while (original->next != NULL || reverse->next != NULL)
{
if (original->data != reverse->data)
return false;
cout << "debug 2" << endl;
original = original->next;
reverse = reverse->next;
}
return true;
}
// #include "solution.h"
node *takeinput()
{
int data;
cin >> data;
node *head = NULL, *tail = NULL;
while (data != -1)
{
node *newnode = new node(data);
if (head == NULL)
{
head = newnode;
tail = newnode;
}
else
{
tail->next = newnode;
tail = newnode;
}
cin >> data;
}
return head;
}
void print(node *head)
{
node *temp = head;
while (temp != NULL)
{
cout << temp->data << " ";
temp = temp->next;
}
cout << endl;
}
int main()
{
node *head = takeinput();
node *revese2 = createReversedLinkedList(head);
print(revese2);
// bool ans = check_palindrome(head);
// if (ans)
// cout << "true";
// else
// cout << "false";
// return 0;
}
As asked by the OP, building a reversed linked is simply done by building as you would a stack (e.g LIFO) rather than duplicating the same original forward chain. For example:
node *createReversedLinkedList(const node *head)
{
node *newHead = NULL;
for (; head; head = head->next)
{
node *p = new node(head->data)
p->next = newHead;
newHead = p;
}
return newHead;
}
Note we're not hanging our copied nodes on the tail of the new list; they're hanging on the head of the new list, and becoming the new head with each addition. That's it. There is no need to craft an identical list, then reverse it; you can reverse it while building the copy to begin with.
A note on the remainder of your code. You have a dreadful memory leak, even if you fix the reversal generation as I've shown above. In your check_palindrome function, you never free the dynamic reversed copy (and in fact, you can't because you discard the original pointer referring to its head after the first traversal:
bool check_palindrome(node *head)
{
node *original = head;
node *reverse = returnReverseLinkedList(head); // only reference to reversed copy
while (original->next != NULL || reverse->next != NULL)
{
if (original->data != reverse->data)
return false; // completely leaked entire reversed copy
original = original->next;
reverse = reverse->next; // lost original list head
}
return true;
}
The most obvious method for combating that dreadful leak is to remember the original list and use a different pointer to iterate, and don't leave the function until the copy is freed.
bool check_palindrome(const node *head)
{
bool result = true;
node *reverse = returnReverseLinkedList(head);
for (node *p = reverse; p; p = p->next, head = head->next)
{
if (p->data != head->data)
{
result = false;
break;
}
}
while (reverse)
{
node *tmp = reverse;
reverse = reverse->next;
delete tmp;
}
return result;
}
class node{
public:
int data;
node *left;
node *right;
};
class BFStree{
public:
void insert(int key);
void deleteNode(int key);
void inorderTraversal(node *temp);
void inorder();
node *root;
BFStree(){
root = NULL;
}
};
void BFStree::insert(int key){
node *temp = root;
while(temp!=NULL){
if(key>temp->data){
temp = temp->right;
}
else if(key<temp->data){
temp = temp->left;
}
else{
cout << "NOT ALLOWED TO HAVE SAME DATA" << temp->data << " " << key << endl;
}
}
node *temp2 = new node;
temp2->data = key;
cout << key << " inserted" << endl;
temp2->left = NULL;
temp2->right = NULL;
temp = temp2;
}
int main(){
BFStree t;
t.insert(7);
t.insert(3);
t.insert(21);
}
I am trying to use the above function to insert data into the bst tree but it doesn't do anything, even the root is NULL
but when I use the following function, the job is accomplished. I don't understand what I have done differently in the two codes. I am new to programming so I am having such bad doubts. Please help me clear it.
void BFStree::insert(int key){
node *temp = root;
if(temp==NULL){
temp = new node;
temp->data = key;
temp->left = NULL;
temp->right = NULL;
root = temp;
return;
}
while(1){
if(key>temp->data){
if(temp->right==NULL){
node *temp2 = new node;
temp2->data = key;
temp2->left = NULL;
temp2->right = NULL;
temp->right = temp2;
break;
}
else{
temp = temp->right;
}
}
else if(key<temp->data){
if(temp->left==NULL){
node *temp2 = new node;
temp2->data = key;
temp2->left = NULL;
temp2->right = NULL;
temp->left = temp2;
break;
}
else{
temp = temp->left;
}
}
}
}
The main problem in your first function is that you just iterate over the tree until you find a null, and after that you assign that null value to temp.
After that you create a new node and assign temp the reference of new node. But your temp node is not connected to the tree. And there is no connection between the temp node and its parent or root node of the tree.
Whereas in 2nd example :
if(temp->right==NULL){
node *temp2 = new node;
temp2->data = key;
temp2->left = NULL;
temp2->right = NULL;
temp->right = temp2;
break;
}
This is the key, you store the reference of your newly created node in the right or left child of its parent node.
I am not able to figure out, why is my code to insert into a sorted doubly linked list failing on some test cases.Please let me know. I dont know of the test cases, they are system generated.
Node* SortedInsert(Node *head,int data)
{
// Complete this function
// Do not write the main method.
Node * temp = (Node*)malloc(sizeof(Node));
temp->data = data;
temp->next = NULL;
temp->prev = NULL;
if (head == NULL)
{
head = temp;
return head;
}
if (temp->data <= head->data)
{
temp->next = head;
head->prev = temp;
head = temp;
return head;
}
Node *curr = head;
while (curr->next!=NULL)
{
if (temp->data <= curr->data)
{
curr->prev->next = temp;
temp->prev = curr->prev;
temp->next = curr;
curr->prev = temp;
return head;
}
curr = curr->next;
}
curr->next = temp;
temp->prev = curr;
return head;
}
Thanks
Once you reach the last node, you should again compare its data with the new node and insert accordingly.
curr->next = temp;
temp->prev = curr;
return head;
}
If execution reaches this part, then at present curr is pointing to the last node. Now you should again compare its data like the following:
if (temp->data <= curr->data)
{ // insert before last node
curr->prev->next = temp;
temp->prev = curr->prev;
temp->next = curr;
curr->prev = temp;
return head;
}
// else insert at the end.
curr->next = temp;
temp->prev = curr;
return head;
}
Alternatively, you can write an if condition for node at the end of the list
Node* SortedInsert(Node *head,int data)
{
struct Node* p = head;
struct Node* q = NULL;
struct Node* r = new Node;
r->data=data;
r->prev=NULL;
r->next=NULL;
if(p==NULL){
p=r;
head=p;
}
else if(p!=NULL&&p->data>r->data){
p->prev=r;
r->next=p;
p->next=NULL;
head = r;
}
else{
p=head;
while(p!=NULL) {
if(p!=NULL&&p->data>r->data){ //If node is not at the end of list and smaller than some node
(p->prev)->next = r;
r->next = p;
r->prev = p->prev;
p->prev=r;
return head;
}
else if(p->next==NULL) //If node is at the end of list
{
p->next = r;
r->prev = p;
r->next = NULL;
return head;
}
else{}
p=p->next;
}
}
return head;
}
/*
* For reference:
*
* link to the problem :-
* https://www.hackerrank.com/challenges/insert-a-node-into-a-sorted-doubly-linked-list
*
* DoublyLinkedListNode {
* int data;
* DoublyLinkedListNode* next;
* DoublyLinkedListNode* prev;
* };
*
* Function to insert a node in sorted DLL with given data & return head
*/
DoublyLinkedListNode* sortedInsert(DoublyLinkedListNode* head, int data) {
DoublyLinkedListNode** prev = &head;
DoublyLinkedListNode* newNode = new DoublyLinkedListNode(data);
while ( *prev && ((*prev)->data <= data) ) {
prev = &(*prev)->next;
}
newNode->prev = *prev;
newNode->next = *prev;
*prev = newNode;
return head;
}
This solution works in all cases (insert in between, insert at the head, insert at the end etc). I was just trying to write a code without any conditions. While searching for some error, I ended up on this SO thread. Do correct me if this solution doesn't work for some test cases.
This is my code that passed all the test cases.
DoublyLinkedListNode* sortedInsert(DoublyLinkedListNode* head, int data) {
DoublyLinkedListNode *temp = head;
if (!temp) {
temp = new DoublyLinkedListNode(data);
head = temp;
return head;
}
while (temp->data < data && temp->next != nullptr) {
temp = temp->next;
}
if (temp->data > data) {
DoublyLinkedListNode *t = new DoublyLinkedListNode(data);
t->next = temp;
if (temp == head) {
head = t;
temp->prev = t;
return head;
}
temp->prev->next = t;
t->next = temp;
temp->prev = t;
} else {
DoublyLinkedListNode *t = new DoublyLinkedListNode(data);
t->next = temp->next;
temp->next = t;
temp->next->prev = temp;
if (t->next != nullptr) t->next->prev = t;
}
return head;
}
Here is some code in order to Insert a Node in Sorted Doubly Linked List
Node* SortedInsert(Node *head,int data)
{
struct Node *temp;
struct Node *newNode = new Node();
newNode->data=data;
newNode->next=NULL;
newNode->prev=NULL;
temp=head;
if(head==NULL)
{
head = newNode;
}
else
{
while(temp!=NULL)
{
if(temp->data<data&&temp->next!=NULL)
{
temp=temp->next;
}
else
{
if(temp->next==NULL&&temp->data<data)
{
temp->next = newNode;
newNode->prev = temp;
break;
}
else if(temp->prev==NULL&&temp->data>data)
{
newNode->next=temp;
temp->prev = newNode;
head=newNode;
break;
}
else
{
newNode->next = temp;
newNode->prev = temp->prev;
temp->prev->next =newNode;
temp->prev=newNode;
break;
}
}
}
}
return head;
}
Here is the complete C++ program for inserting a node at appropriate position in a sorted doubly linked list:
void InsertInSortedDoublyLinkedList()
{
Node *head;
head = CreateDoublyLinkList(3);
PrintDoublyLinkedList(head);
head = SortedInsert(head, 6);
printf("\n Linked list after insertion in sorted order\n");
PrintDoublyLinkedList(head);
}
void PrintDoublyLinkedList(Node *head)
{
Node *temp = head;
printf("NULL -> ");
while (temp)
{
printf("%d ->", temp->data);
temp = temp->next;
}
printf("NULL");
}
Node* CreateDoublyLinkList(int numberOfNodes)
{
Node *head = NULL, *temp, *temp1;
int startingValue = 3;
if (numberOfNodes == 0)
{
return head;
}
temp = (Node*)malloc(sizeof(Node));
temp->data = startingValue;
temp->next = NULL;
temp->prev = NULL;
head = temp;
startingValue += 2;
numberOfNodes--;
for (; numberOfNodes > 0; numberOfNodes--, startingValue += 2, temp = temp->next)
{
temp1 = (Node*)malloc(sizeof(Node));
temp1->data = startingValue;
temp1->next = NULL;
temp1->prev = temp;
temp->next = temp1;
}
return head;
}
Node* SortedInsert(Node *head,int data)
{
Node *temp = NULL, *temp1,*newNode;
int nodeInserted = 0;
if (head == NULL)
{
head = (Node*) malloc(sizeof(Node));
head->data = data;
head->next = NULL;
head->prev = NULL;
}
else
{
if (head->data > data)
{
//insertion need to take place before head itself.
newNode = (Node*)malloc(sizeof(Node));
newNode->data = data;
newNode->next = head;
newNode->prev = NULL;
head->prev = newNode;
head = newNode;
}
else
{
temp1 = head;
temp = head ->next;
while (temp)
{
if (temp->data > data)
{
//we need to insert the node before temp
newNode = (Node*)malloc(sizeof(Node));
newNode->data = data;
newNode->prev = temp1;
newNode->next = temp;
temp1->next = newNode;
temp->prev = newNode;
nodeInserted = 1;
break;
}
temp1 = temp;
temp = temp->next;
}
if (!nodeInserted)
{
//node insertion need to take place at tail.
newNode = (Node*)malloc(sizeof(Node));
newNode->data = data;
newNode->prev = temp1;
newNode->next = NULL;
temp1->next = newNode;
}
}
}
return head;
}
struct Node
{
int data;
Node *next;
Node *prev;
}
Node* SortedInsert(Node *head,int data)
{
Node* p1,*p2;
int n=10;
p2=(Node*)malloc(sizeof(struct Node));
p2->next=NULL;
p2->prev=NULL;
p2->data=data;
p1=head;
if(p1==NULL)
return p2;
while(p1->next!=NULL)
{
if(p1->data<data)
p1=p1->next;
else
break;
}
/*Three cases arise when p1->next == NUll i.e. we are end of list
case 1: insert after the end
case 2: insert in between last and second last node
case 3: insert node at the beginning of the list
*/
if(p1->next==NULL)
{
if(p1->data<data)
{
p2->prev=p1;
p1->next=p2;
}
else if(p1->data > data && p1->prev!=NULL)
{
p2->next=p1;
p2->prev=p1->prev;
if(p1->prev!=NULL)
p1->prev->next=p2;
p1->prev=p2;
}
else
{
p2->next=p1;
p2->prev=p1->prev;
if(p1->prev!=NULL)
p1->prev->next=p2;
p1->prev=p2;
head=p2;
}
}
// Here we have only one case where new node is inserted between two nodes of the list
else
{
p2->next=p1;
p2->prev=p1->prev;
if(p1->prev!=NULL)
p1->prev->next=p2;
p1->prev=p2;
}
return head;
}