We can print the linked list in reverse easily with the help of recursion
void reverseDisplay(nodeType* head)
{
if( head == NULL) // Check if the node is the last node
return;
reverseDisplay(head -> next); // Go to the next node
printf("%d \t ", head->data); // Print the data in the node
}
To summarise, the traverse to the end of the list by recursively calling the reverseDisplay() function, and then print the nodes in reverse order
Write a function named reverse Display that takes a pointer to the first node of a...
By definition, the height of a node in a binary tree is the number of edges along the longest path from the node to any leaf. Assume the following node structure struct TreeNode! int data; node Type right; // points to right child node Type "left; // points to left child }; Write a recursive function that takes a pointer to a node in a binary tree and returns its height. Note: the height of a leaf node is o...
2) (10 pts) Write a function that takes in a pointer to a linked list of nodes storing integers and a variable named value, and returns the number of nodes in the list storing that value. For example, if a list pointed to by listPtr stores 2, 6, 2, 3, 4, 2, 6, and 6 and value = 6, your function should return 3, since 6 appears in the list 3 times. Please use the struct and function prototype provided...
Given the node structure and the head pointer (headptr) of a linked list write a code to move the last element to the head of the list. struct node { int value; struct node *next; };
13. Given the following structure struct node { struct node *next; int id; }; Write a code to insert a new node into the linked list as the last node of the linked list struct node *insertLast(stuct node **head, int newId) { } 14. Given the following structure struct node { struct node *next; int id; }; Write a code to insert a new node into the linked list after a node with the same id as the input parameter...
Write a C++ function to add a node to the beginning of a linked list. Your function takes two arguments - the head of the linked list and the value num to be added. Note that the list may be empty! Your function should modify the head of the linked list to point to the new node, and set the new node to point to the rest of the list (if not empty). Example: Initial Array: 4->2->3, key = 5...
Structure struct Node int Manth; // Mont h double dAvg: 1/ Average struct Node pNext // with, the linked İist 3hown above the function will return gven that the average is 3.8 Ptr to next -Node; Ret (3,3.8) (4,2.5) (20pts)( Recursive function) Show the code for a function that receives a pointer to the head of an ordered singly linked list that uses the structure in the top left. The function will return the pointer node that shows the highest...
A linked list is constructed of nodes described by the following structure: struct node{ char data; struct node *next; }; Assume a linked list containing a sentinel node is constructed from the above nodes. Write a function named "count"-prototyped as int count(struct node*sent)- that accepts a pointer to the sentinel node; counts the number of data (non-sentinel) nodes containing the character 'A'; and returns that count as the function value.
given a pointer to the first node of a linked list, you are asked to reverse the list. Explain, in English, not code, how you would complete such a task if you had a limited amount of memory (not enough to store the whole list twice!)
A linked list of integers is built using the following struct: struct node { int data; struct node *next; }; Define a function named max that returns the maximum integer in a list. The function takes one arguments, a pointer to the head of the list. The function returns an integer, which is the maximum value. If the list is empty, return zero. NOTE: You know nothing about the values in the list. They could all be negative!
C++ Consider the following structure of node and linked list. struct Node { int key; Node *next; }; 10 -> 20 -> 30 -> 10 -> 10 -> 50 -> 10 -> NULL What will be the output of following pseudo-code? Consider head is the pointer to the first node of above linked list. Node *walker = head; int count = 0; while(walker!= NULL && count < 3) { if(walker->key == 10) { count = count + 1; } walker...