de (Dummy Head) must be existed with a Doubly link

Question 1: Add a member function to LinkedList called insertAfter. The function should insert a new node after another Node that contains a target value. Your function should do nothing if the target value is not in the list. Question 2: Write a function that takes these parameters Node* list1 Node* list2 Concatenate list1 onto list 2 so that they form one long linked list. Question 3: Create a circular Linked list. It has node 1, 2, 3, 4, 5. And the last node points to node 1.

A)Delete Node-3 from the Linked List. B)Insert New Node-3 between Node-2 and Node-4.

(a) Explain the advantages of adding a tail pointer to a linked list.(b) Explain the advantages of doubly-linked over singly-linked lists.

How nodes are defined (struct node (value count left right) #:mutable #:transparent)     Write in Racket (traverse n) A traversal of a BST is an algorithm for “visiting” all node in the BST. The traversal must visit each node exactly once. In the case of a linked list, a traversal is trivial since the structure is linear: start at the head, move to the next node, and stop when you reach the tail. In the case of a BST, traversal must account for multiple child nodes and keep track of which subtrees have already been visited and which have not. There are three types of traversal: in-order, pre-order, and post-order. We will only implement in-order. The in-order traversal of a BST has the property that the node values will display in ascending or sorted order. The function can be defined either recursively or iteratively. Recursion is much simpler, so we’ll stick to that.               Recursive Algorithm for In-Order Traversal of BST parameter: node n, the root of the tree…

Given main(), define an InsertAtEnd() member function in the ItemNode class that adds an element to the end of a linked list. DO NOT print the dummy head node.

} (1 of 3 parts) Consider the function below that is given the head and tail pointers to a double linked list. bool patronum (Node ✶h, Node * t) { bool c = true; if (t nullptr) return true; Node* p = t; while (p != h) { Node* j = p; Node* r = p->prev; EXAM for (Node* r = p->prev; r != nullptr; r = r->prev) { if (j->value > r->value) { c = false; j = r; } swap(p->value, j->value); p = p->prev; return c;

Consider the following structure of a node for a linked list. struct Node { int key; Node *next; }; What will be the output of the following pseudo-code? Assume, head is a pointer to the first node of the following linked list. 5->2->8->2->9->2->11-> NULL Node *pres head; 0; while(pres!= NULL && Count key : int count 2) { count = count + 1; } pres = pres->next; } cout key « endl; 11 Error 2.

Write a function, to be included in an unsorted linked list class, called getLargest, that will return the largest item in the list.