What is the left child and right child of the root afterexecuting the following code binary_Search_Tree bt1-new binar Search_Tree (); bt1.insert(50): bt1.insert(90); bt1.insert(75): bt1.insert(82); bt1.insert(66); bt1.insert(69); bt1.insert(100): bt1.delete(66): bt1.insert(17); bt1.insert(23); bt1.insert(35); bt1.delete(17) bt1.insert(47): bt1.delete(90); O (3 Points) left = 17, right 90 O left = 17, right 100 left = 17, right 75 O left 23, right= 100 No Corect Answer

What is the left child and right child of the root afterexecuting the following code binary_Search_Tree bt1-new binar Search_Tree (); bt1.insert(50): bt1.insert(90); bt1.insert(75): bt1.insert(82); bt1.insert(66); bt1.insert(69); bt1.insert(100): bt1.delete(66): bt1.insert(17); bt1.insert(23); bt1.insert(35); bt1.delete(17) bt1.insert(47): bt1.delete(90); O (3 Points) left = 17, right 90 O left = 17, right 100 left = 17, right 75 O left 23, right= 100 No Corect Answer

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

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def remove_after_kth(items, k=1): Given a list of items, some of which may be duplicated, create and return a new list that is otherwise the same as items, but only up to k occurrences of each element are kept, and all occurrences of that element after those first k are discarded. Hint: loop through the items, maintaining a dictionary that remembers how many times you have already seen each element. Update this count as you go, and append each element to the resultlist only if its count is still at most equal to k. items k expected results [42, 42, 42, 42, 42, 42, 42] 3 [42, 42, 42] ['tom', 42, 'bob', 'bob', 99, 'bob', 'tom', 'tom', 99] 2 ['tom', 42, 'bob', 'bob', 99, 'tom', 99] [1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1] 1 [1, 2, 3, 4, 5] [1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5] 3 [1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 1, 5] [42, 42, 42, 99, 99, 17] 0 [ ] Please add comments between the codes so I can understand
Do it asap. I will helpfup.
Inserting Nodes into the List In the following code (This is Code By C) : there are 3 functions : display, insert_begin, and delete. The request is add two functions : First Function (insert_last) : to insert values from last. Second Function (insert_mid): to insert values from Middle. The code: #include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h> struct node{int info;struct node* next;}; struct node* start; void display();void insert_begin(); void delete(); int main(){ int choice;start = NULL; while (1){printf("\n\t\t MENU \t\t\n");printf("\n1.Display\n");printf("\n2.Insert at the beginning\n");printf("\n3.Insert at the end\n");printf("\n4.Delete\n");printf("\n5.Exit\n"); printf("Enter your choice:\t");scanf_s("%d", &choice); switch (choice){case 1:display();break;case 2:insert_begin();break;case 3:delete();break;case 4:exit(0);break;default:printf("\nWrong Choice.\n");break;} }return 0;} void display(){struct node* temp;if (start ==…
Complete explanation is required with code Don't do incomplete please
C++ please Please complete Programming Exercise 4, from pages 1402, from chapter 19 from your textbook. Write a function, singleParent, that returns the number of nodes in abinary tree that have only one child. Add this function to the classbinaryTreeType and create a program to test this function. (Note: Firstcreate a binary search tree.)
Computer Science After you run the program and understand it's logic, modify word_frequency.py to use a binary search tree instead of a dictionary. All you need to do is to import the binary search tree, and change one line of code in the program! ---word_frequency.py--- import sysfilename = sys.argv[1] freq = {}for piece in open(filename).read().lower().split():# only consider alphabetic characters within this pieceword = ''.join(c for c in piece if c.isalpha())if word: # require at least one alphabetic characterfreq[word] = 1 + freq.get(word, 0) max_word = ''max_count = 0for (w,c) in freq.items(): # (key, value) tuples represent (word, count)if c > max_count:max_word = wmax_count = cprint('The most frequent word is', max_word)print('Its number of occurrences is', max_count)
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def animals(animal_dict, target): """ Filter animals based on the target category. :param animal_dict: A dictionary mapping animals to their categories. :type animal_dict: dict :param target: The target category to filter animals. :type target: str :return: A list of animals belonging to the target category. :rtype: list animals_dict = {"dogs": "mammals", "snakes": "reptiles", \ "dolphins": "mammals", "sharks": "fish"} >>> target = "mammals" >>> animals(animals_dict, target) ['dogs', 'dolphins'] >>> animals_dict = {True: "mammals"} >>> animals(animals_dict, target) # This will raise an AssertionError Traceback (most recent call last): ... AssertionError: Target should be provided and not None """ assert target is not None, "Target should be provided and not None" assert isinstance(animal_dict, dict) assert all(isinstance(animal, str) for animal in animal_dict.keys())…
Which of the following data structures can we use the [] operator on? Arrays B-Trees Maps Multimaps
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Which of the following is true about Set objects? It enables efficient search of information Removal of elements are possible by moving other elements Each member of a Set object is associated with an index Set member reveal the order of insertion of items