Define a function called get_predecessor() which takes a Node object and a value as parameters. The function should search the chain of nodes starting at the parameter node and returns the node which comes before the node containing the parameter value. If the parameter value is NOT in the chain of nodes or the parameter value is in the first node of the chain, the function returns None. Test: x = Node('hello') y Node('world') z = Node('goodbye') x.set_next(y) y.set_next(z) print(get_predecessor(x, "goodbye')) Output: world Test: x = Node('hello') y = Node('world') z = Node('goodbye') x.set_next(y) y.set_next(z) print(get_predecessor(x, 'hello')) Ouput: None

Define a function called get_predecessor() which takes a Node object and a value as parameters. The function should search the chain of nodes starting at the parameter node and returns the node which comes before the node containing the parameter value. If the parameter value is NOT in the chain of nodes or the parameter value is in the first node of the chain, the function returns None. Test: x = Node('hello') y Node('world') z = Node('goodbye') x.set_next(y) y.set_next(z) print(get_predecessor(x, "goodbye')) Output: world Test: x = Node('hello') y = Node('world') z = Node('goodbye') x.set_next(y) y.set_next(z) print(get_predecessor(x, 'hello')) Ouput: None

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

See similar textbooks

Similar questions

A BST is constructed in the usual way using the node definition below. Write a function int child2( bst node t *curr) that returns the number of nodes that have 2 children. typedef struct BST NODE T {int data;struct BST NODE T *left, *right; } bst node t; int child2( bst node t *curr){ }
Consider the following function:int mystery(NodeInt32* node){int counter = 0; while (node != NULL) {counter++;node = node->next; } return counter;}
Write a c++ code of link list which is used to take your data like name, mob, age , marks. After that you will display the data in Disp() function. Your code will be able to take 5 students data and also check which student marks is greater than from other. At the end you will delete any node from link list Please should create grading_fun(); Bonus > 80 A 70 B 60 C 50 D
In C++ Question A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example: head refers to the list of nodes 1 → 2 → 3 → 1 → NULL. There is a cycle where node 3 points back to node 1, so return 1. Function Description: Complete the has cycle function provided in the moodle assignment folder. It has the following parameter: • SinglyLinkedListNode pointer head: a reference to the head of the list. Returns: • int: 1 if there is a cycle or 0 if there is not. Note: if the list is empty, head will be null. Input Format: The code stub reads from stdin and passes the appropriate argument to your function. The custom test cases format will not be described for this question due to its complexity. Expand the section for the main function and review the code if you would like to figure out how to create…
Modify the given code to accept float instead of int as a coefficient in C LANGUAGE: // Node structure containing power and coefficient of// variablestruct Node {int coeff;int pow;struct Node* next;}; // Function to create new nodevoid create_node(int x, int y, struct Node** temp){struct Node *r, *z;z = *temp;if (z == NULL) {r = (struct Node*)malloc(sizeof(struct Node));r->coeff = x;r->pow = y;*temp = r;r->next = (struct Node*)malloc(sizeof(struct Node));r = r->next;r->next = NULL;}else {r->coeff = x;r->pow = y;r->next = (struct Node*)malloc(sizeof(struct Node));r = r->next;r->next = NULL;}} // Function Adding two polynomial numbersvoid polyadd(struct Node* poly1, struct Node* poly2,struct Node* poly){while (poly1->next && poly2->next) {// If power of 1st polynomial is greater then 2nd,// then store 1st as it is and move its pointerif (poly1->pow > poly2->pow) {poly->pow = poly1->pow;poly->coeff = poly1->coeff;poly1 =…
Concatenate Map This function will be given a single parameter known as the Map List. The Map List is a list of maps. Your job is to combine all the maps found in the map list into a single map and return it. There are two rules for addingvalues to the map. You must add key-value pairs to the map in the same order they are found in the Map List. If the key already exists, it cannot be overwritten. In other words, if two or more maps have the same key, the key to be added cannot be overwritten by the subsequent maps. Signature: public static HashMap<String, Integer> concatenateMap(ArrayList<HashMap<String, Integer>> mapList) Example: INPUT: [{b=55, t=20, f=26, n=87, o=93}, {s=95, f=9, n=11, o=71}, {f=89, n=82, o=29}]OUTPUT: {b=55, s=95, t=20, f=26, n=87, o=93} INPUT: [{v=2, f=80, z=43, k=90, n=43}, {d=41, f=98, y=39, n=83}, {d=12, v=61, y=44, n=30}]OUTPUT: {d=41, v=2, f=80, y=39, z=43, k=90, n=43} INPUT: [{p=79, b=10, g=28, h=21, z=62}, {p=5, g=87, h=38}, {p=29,…
PYTHON: Please fix my code to print the correct traversals, especially at the def functions, because I could not print the correct results of the inorder, postorder, and preorder traversals from the user input. #POSTORDERclass Node: def __init__(self, data): self.left = None self.right = None self.data = data # Insert Node def insert(self, data): if self.data: if data < self.data: if self.left is None: self.left = Node(data) else: self.left.insert(data) elif data > self.data: if self.right is None: self.right = Node(data) else: self.right.insert(data) else: self.data = data # Print the Tree def PrintTree(self): if self.left: self.left.PrintTree() print(self.data),…
Maze Runner Function - Implementation of this function is done in a1_partd.py We describe a maze as having row x col cells. For example if row was 3, and col was 4, then we would have a grid of cells as follows. We describe a wall by the two cell numbers the wall separates. If every single wall existed, there would be (row-1)(col) + (col-1)(row) walls. 0 | 1 | 2 | 3 4 | 5 | 6 | 7 8 | 9 | 10 | 11 A Maze class (which you do not need to implement) describes a maze as mentioned above. This class is defined in maze.py. It has methods that you can use to travel through the maze (i.e. figure out where you are, find a neighbour cell etc.) use a recursive maze runner function: def find_path(maze, from_cell, to_cell); The find_path function will find a path from cell number from_cell to cell number to_cell and will return it as a list containing all the cell numbers along the path, from the from_cell to the to_cell. You are allowed to use this function as a wrapper to a recursive function that…
Write a function called oneChild, that returns the number of nodes with exactly one child. Hint just the function
Write a function that takes two queue structures as parameters: original queue and modified queue. The function finds the numbers greater than or equal to 5 in the original queue and places 1 to the same position in the modified queue. For the numbers less than 5, modified queue element has 0. Assume that two input queues have the same number of elements at a time. Example Original Queue: 351628 Example Modified Queue: 01101 void modifyqueue (struct queue *org, struct queue *mod){ Note that you need printqueue function to display the queue contents in order to verify your solutions.
Create a node class named LinkedNodes that uses up to 4 dynamic pointers to connect it to the other 8 nodes in the structure (a total of 9). The data in each node will be a unique integer number from 1 through 9. Include in your class definition these functions at a minimum: getVal(), setVal(), getNext(), setNext(), getPrev(), setPrev(), constructor(s) and destructor. Write a simple program to load these nodes into a linked structure. Do not use an array, the pointers will do that for you. You should be able to reach any node from any other node in the structure by traversing a maximum of 2 nodes from the current one. The output from your program should report the traversals between all nodes in the structure, starting with the node whose value is one. (See the example below) Node 1 2 3 4 5 6 7 8 9 -> Traverses to Node 1 Node 2 1->2 2->6->1 3->6->1 3->7->2 4->6->1 4->7->2 4->8->3 5->6->1 5->7->2 5->8->3 6->1->2 6->1->3 7->1->3 8->3 6->1 7->1 8->1 9->1 Node 3 1->3 2->3 7->2 8->2 9->2…
Write a driver function called size that takes as its parameter a linked list (nodeType pointer). The function should count how many nodes are in the list and return that value to the calling function.