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How can a binary search tree insert a new node at the root position?
To insert a new node at the root position in a binary search tree, the tree must be restructured by following these steps: Create a new node with the desired value. Compare the value of the new node with the value of the current root node. If the new node's value is less than the root node's value, set the left child of the root node to be the current root node, and set the left child of the new node to be the previous left child of the root node. If the new node's value is greater than the root node's value, set the right child of the root node to be the current root node, and set the right child of the new node to be the previous right child of the root node. Set the new node as the new root of the binary search tree. By following these steps, a new node can be inserted at the root position of a binary search tree while maintaining the binary search tree properties.
How does binary tree insertion work and what are the key steps involved in inserting a new node into a binary tree?
Binary tree insertion involves adding a new node to a binary tree while maintaining the tree's structure. The key steps in inserting a new node are: Start at the root node and compare the value of the new node with the current node. If the new node's value is less than the current node, move to the left child node. If it is greater, move to the right child node. Repeat this process until reaching a leaf node (a node with no children). Insert the new node as the left or right child of the leaf node, depending on its value compared to the leaf node's value.
Can structure contain a pointer itself?
Yes a simple exp is the link list. struct node { int data; struct node *link; }
How can a binary tree be converted into a doubly linked list?
To convert a binary tree into a doubly linked list, perform an in-order traversal of the tree and adjust the pointers to create the doubly linked list. This involves setting the left child pointer to the previous node and the right child pointer to the next node in the list.
How can I implement a merge sort algorithm for a doubly linked list in Java?
To implement a merge sort algorithm for a doubly linked list in Java, you can follow these steps: Divide the doubly linked list into two halves. Recursively sort each half using merge sort. Merge the two sorted halves back together in sorted order. You can achieve this by creating a mergeSort() method that takes the doubly linked list as input and recursively divides and merges the list. Make sure to handle the merging process for doubly linked lists by adjusting the pointers accordingly. Here is a basic outline of how you can implement this algorithm in Java: java public class MergeSortDoublyLinkedList public Node mergeSort(Node head) if (head null head.next null) return head; Node middle getMiddle(head); Node nextOfMiddle middle.next; middle.next null; Node left mergeSort(head); Node right mergeSort(nextOfMiddle); return merge(left, right); private Node merge(Node left, Node right) if (left null) return right; if (right null) return left; Node result null; if (left.data right.data) result left; result.next merge(left.next, right); result.next.prev result; else result right; result.next merge(left, right.next); result.next.prev result; return result; private Node getMiddle(Node head) if (head null) return head; Node slow head; Node fast head; while (fast.next ! null fast.next.next ! null) slow slow.next; fast fast.next.next; return slow; class Node int data; Node prev; Node next; public Node(int data) this.data data; This code snippet provides a basic implementation of the merge sort algorithm for a doubly linked list in Java. You can further customize and optimize it based on your specific requirements.
How would you in C programming using a Node struct and a Linkedlist struct take user integer inputs and store them in the linked list in numerical order and then print them in order?
#include<stdio.h> #include<malloc.h> typedef struct linkedlist_node_t { int value; linkedlist_node_t* next; } linkedlist_node; typedef struct linkedlist_t { linkedlist_node* head; } linkedlist; /* Deletes all nodes in the given list. */ void delete_all (linkedlist* list) { /* Ensure the list is non-null. */ if (!list) return; linkedlist_node* old_head; while (list->head) { /* Copy the head pointer. */ old_head = list->head; /* Make its next node the new head. */ list->head = old_head->next; /* Release the old head*/ free (old_head); } /* The list is empty. */ list->head = 0; } /* Inserts the given value into the given linked list. */ linkedlist_node* insert_value (linkedlist* list, int value) { /* Ensure the list is non-null. */ if (!list) return 0; /* Instantiate and initialise a new node. */ linkedlist_node* new_node = malloc (sizeof (linkedlist_node)); new_node->value = value; new_node->next = 0; if (list->head==0) { /* If the list is empty (has no head), the new node becomes the head. */ list->head = new_node; } else if (value<list->head->value) { /* Otherwise, if the value is less than the head's value, insert the new node before the head node. The new node becomes the head. */ new_node->next = list->head; list->head = new_node; } else { /* Otherwise, traverse the list to find the insertion point (the node that comes before the new node). */ linkedlist_node* node = list->head; while (node->next && node->next->value<value) node = node->next; /* Insert the new node immediately after node. */ new_node->next = node->next; node->next = new_node; } /* Return the new node (not essential, but good practice). */ return new_node; } /* Prints the given list. */ void print_list (linkedlist* list) { /* Ensure the list is non-null. */ if (!list) return; /* Traverse the list from the head, printing each node on a new line. */ linkedlist_node* n = list->head; while (n) { printf ("%d\n", n->value); n = n->next; } } /* Test-drive the functions. */ int main() { /* Instantiate and initialise an empty list (no head). */ linkedlist list; list.head = 0; /* Insert values (in descending order). */ for (int v=0; v<10; ++v) insert_value (&list, 10-v); /* Print the list (in ascending order). */ print_list (&list); /* Don't forget to release resources! */ delete_all (&list); return 0; }
Write algorithm to delete last node in circular linked list?
This depends on whether the list is singly or doubly(or multiply) linked, and on the actual implementation of the list. For example, you can write a CDLL(circular doubly linked list) without maintaining your beginning or ending nodes, using only a current pointer, thus this question doesn't really apply as there would be no "last" node and thus it would be like deleting any node.A typical implementation of a circular singly-linked list (CSLL) list actually maintains the pointer to the last element (hence it's FIFO nature) and thus there are both last and first nodes.This deletion is a little tricky. Consider that you have situations where the next pointer will point to the current element. On the other hand, you also have a situation where there are n-values that you have to iterate over to find the next-to-last value. Typically you would delete the first node in these lists, again dictated by the FIFO nature of these lists, but deletion of the last node is also not impossible.set struct node *last to list->endif (list->end->next == list->end){set list->end to null (leaving an empty list)} else {while(true){if(last->next == list->end){break}set last to last->next}set link->last to list->end->next (this temporarily sets list's end node to current first node)free last->next (frees the last node)set last->next to list->end (set the new last node next pointer to the first node)set list->end to last (set the list's end node to the new last node)}
How do you write a C program to add a node in a linked list after a given node?
Given the 'given node' G, after which the 'new node' N is to be added: 1. For a single-linked list, where the node structure contains a pointer to the next node in the list (or a dedicate invalid pointer value, such as NULL) in a field named 'next', you set N.next = G.next; G.next = &N; 2. For a double-linked list of the same characteristic as above, but with an added 'prev' pointer to point to the previous node, you'd set N.next = G.next; G.next = &N; N.prev = &G;if (N.next) N.next.prev = &N;
How can i find last node of a cicular list whose size i don't know and the last node points to any other node except first node of the link list through c plus plus language of data structure?
How can i find last node of a cicular list whose size i don't know and the last node points to any other node except first node of the link list through c plus plus language of data stucture?
What is the advantage of pointing last node instead of start in circular link list?
This is only an advantage in singly-linked lists as it allows O(1) constant time access to both the last node and the first node because the first node is always the last node's next node. If you only pointed to the first node then you'd have O(1) constant time access to the first node, but the last node would require a complete traversal of the list, which takes O(n) linear time for a list with n nodes.
Algoritm for deleting the last element from a list?
Given a list and a node to delete, use the following algorithm: // Are we deleting the head node? if (node == list.head) { // Yes -- assign its next node as the new head list.head = node.next } else // The node is not the head node { // Point to the head node prev = list.head // Traverse the list to locate the node that comes immediately before the one we want to delete while (prev.next != node) { prev = prev.next; } end while // Assign the node's next node to the previous node's next node prev.next = node.next; } end if // Before deleting the node, reset its next node node.next = null; // Now delete the node. delete node;
How do you insert a node using doubly linked list in c plus plus?
#include<iostream> #include<list> void print_list (std::list<int>& list) { for (std::list<int>::const_iterator it=list.begin(); it!=list.end(); ++it) std::cout << *it << ", "; std::cout << "\b\b \n"; } int main() { // instantiate a list std::list<int> list; // push a new value onto the back of the list: list.push_back (1); print_list (list); // push a new value onto the front of the list: list.push_front (2); print_list (list); // insert a new value at the back of the list list.insert (list.end(), 3); print_list (list); // insert a new value at the front of the list list.insert (list.begin(), 4); print_list (list); // locate value 1. std::list<int>::const_iterator it=list.begin(); while (it!=list.end() && *it != 1) ++it; // insert a new value in front of value 1. list.insert (it, 5); print_list (list); } Output: 1 2, 1 2, 1, 3 4, 2, 1, 3 4, 2, 5, 1, 3
What is cprogram code to insert a node in single link list?
Write Code to Insert a Node in a Single Linked List at any given Position.
Difference between circular and single linklist?
A regular linked list will have a pointer to the start of the list, with each node pointing to the next node, and finally the last node points to NULL. In a circular linked-link, the last node will point to the first node, making the list circular. This requires extra checks to ensure that you don't end up going into an infinite loop while traversing the list.
How many pointers will have to be changed if a node is deleted from a linear linked list?
For a singly-linked list, only one pointer must be changed. If the node about to be deleted (let's call it node for the sake of argument) is the head of the list, then the head node pointer must be changed to node->next. Otherwise, the node that comes before the deleted node must change its next pointer to node->next. Note that given a singly-linked node has no knowledge of its previous node, we must traverse the list from the head in order to locate that particular node, unless the node is the head of the list: void remove (List* list, Node* node) { if (!list !node) return; // sanity check!if (list->head == node) {list->head = node->next;} else {Node* prev = list->head;while (prev->next != node) prev = prev->next; // locate the node's previous nodeprev->next = node->next;}} Note that the remove function only removes the node from the list, it does not delete it. This allows us to restore the node to its original position, because the node itself was never modified (and thus still refers to its next node in the list). So long as we restore all removed nodes in the reverse order they were removed, we can easily restore the list. In order to delete a node completely, we simply remove it and then free it:void delete (List* list, Node* node) {if (!list !node) return; // sanity check!remove (list, node);free (node);} For a doubly-linked list, either two or four pointers must be changed. If the node about to be deleted is the head node, then the head node pointer must be changed to n->next and n->next->prev must be changed to NULL, otherwise, n->prev->next becomes n->next. In addition, if the node about to be deleted is the tail node, then the tail node pointer must be changed to n->prev and n->prev->next must be changed to NULL, otherwise, n->next->prev becomes n->prev. Deletion from a doubly-linked list is generally quicker than deletion from a singly linked list because a node in a doubly-linked list knows both its previous node and its next node, so there's no need to traverse the list to locate the previous node to the one being deleted. void remove (List* list, Node* node) {if (!list !node) return; // sanity check!if (list->head == node) {list->head = node->next;node->next->prev = NULL;} else {node->prev->next = node->next; }if (list->tail == node) {list->tail = node->prev;node->prev->next = NULL;} else {node->next->prev = node->prev; }} Again, to physically delete the node we simply remove and then free the node:void delete (List* list, Node* node) {if (!list !node) return; // sanity check!remove (list, node); free (node); }
How can you insert a new node in a linked list before any given node within the linked list in O of 1?
you dont. you insert a node after it and copy all contents of the previous node in it and put the new values in the old node. just make sure to keep the connections intact.
How do you find out the first node in a list?
The first node in a list is something that you need to keep track of explicitly. While it's possible to find any other node given the first, you must have a starting point stored somewhere.
