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?

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?

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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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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.
Binary tree sort employing a self-balancing binary search tree takes O(n log n) time in the worst scenario, slower than merge sort.
30. In the implementation for a breadth-first search we studied, a queue was used. The code below replaces the queue with a stack. List the pre-order enumeration that the vertices in the graph below are visited using this modified method, starting from vertex 0. } 3 /** stack-based search */ static void modSearch (Graph G, int start) { Stack S new AStack (G.n()); S.push(start); } 6 G.setMark (start, VISITED); while (S.length) > 0) { } modSearch: intv S.pop(); PreVisit (G, v); for (int w = G.first (v); w < G.n(); w = G.next(v, w)) if (G.getMark (w) == UNVISITED) { G.setMark (w, VISITED); S.push(w);
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.