Answered: Even in the worst case, a…

Even in the worst case, a self-balancing binary search tree takes O(n log n) time to sort a binary tree, substantially longer than merge sort.

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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Concept explainers

Merge sort

Merge Sort recursively sorts a given collection of numbers or elements by using the divide and conquer strategy in less time. It is one of the most used algorithms, with the worst-case time complexity of O (n log n). It was invented by John Von Neumann. The algorithm continuously divides an array into sub-arrays until each sub-array consists of a single element. It then merges those sub-arrays to produce a sorted array.

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Even in the worst case, a self-balancing binary search tree takes O(n log n) time to sort a binary tree, substantially longer than merge sort.

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This algorithm find the value of the largest item in a binary search tree. What is the time complexity of this algorithm, and why? int maxValue(node *root) { int max; max = maximum(root) return max; // if max equals -1, then the binary search tree was empty } int maximum(node *node) { if (node == NULL) { return -1; { else { int value = node->value; int left = maximum(node->left); int right = maximum(node->right); if (value < left) { value = left; } if (value < right) { value = right; } return value; } {
In the worst-case scenario, binary tree sort employing a self-balancing binary search tree requires O(n log n) time.
Even in the worst case scenario, using a self-balancing binary search tree to do a binary tree sort takes O(n log n) time, which is much longer than the time required for merge sort.
Using a self-balancing binary search tree, binary tree sort is nevertheless slower than merge sort, taking O(n log n) time in the worst case.
Data Structure and algorithms
Given an arbitrary binary tree T with integer keys stored at the nodes, design an efficient algorithm which determines whether or not T is a binary search tree.
In the worst-case scenario, binary tree sort employing a self-balancing binary search tree requires O(n log n) time, which is slower than merge sort.
Binary tree sort employing a self-balancing binary search tree takes O(n log n) time in the worst scenario, slower than merge sort.
Even when employing a self-balancing binary search tree to construct binary tree sort, the worst-case time required is O(n log n), making it slower than merge sort.
In the worst-case situation, a binary tree sort that employs a self-balancing binary search tree requires temporal complexity of the kind O(n log n). Is this approach much more sluggish than merge sort?
Given a sorted array of n comparable items A, create a binary search tree from the items in A which has height h <= log: n. Your algorithm must create the tree in O(n) time.