The following algorithm is used to recursively count the number of leaves in a binary tree, where T is a binary tree and n is the number of leaves. LeafCounter(T) if (T == EmptySet) then n = 0 else n = LeafCount(T.left)+LeafCount(T.right) return n Is the algorithm above correct? If not, rewrite the pseudocode

Recursion Trees Use the recursion tree method to determine the asymptotic upper bound of T(n). T(n) satisfies the recurrence T(n) = 2T(n-1) + n, and T(0)=0.

Write a recursive form of the following: 0! = 1 n! = n * (n - 1)!

Devise a recursive algorithm for finding n! mod m where n and m are positive integers.

Simple MIPs Recursive Procedure Write a program that calls a recursive procedure. Inside this procedure, add 1 to a counter so you can verify the number of times it executes. Put a number in a MIPS counter that specifies the number of times you want to allow the recursion to continue. You can use a LOOPinstruction (or other conditional statements using MIPS), find a way for therecursive procedure to call itself a fixed number of times.

Using recursion, write a function sum that takes a single argument n and computes the sum of all integers between 0 and n inclusive. Do not write this function using a while or for loop. Assume n is non-negative. def sum(n): """Using recursion, computes the sum of all integers between 1 and n, inclusive. Assume n is positive.>>> sum(1) 1>>> sum(5) # 1 + 2 + 3+ 4+ 5 15 "*** YOUR CODE HERE ***"

Recursive algorithm always gives cleaner code with less cost. Group of answer choices True False

Complete the recursive function remove_last which creates a new list identical to the input list s but with the last element in the sequence that is equal to x removed. Hint: Remember that you can use negative indexing on lists! For example 1st [-1] refers to the last element in a list 1st, lst [-2] refers to the second to last element... def remove_last(x, s): """Create a new list that is identical to s but with the last element from the list that is equal to x removed.>>> remove_last(1, []) []>>> remove_last(1, [1]) []>>> [1]>>> [2] remove_last(1, [1,1]) remove_last(1, [2,1])>>> remove_last(1, [3,1,2]) [3, 2]>>> remove_last(1, [3,1,2,1]) [3, 1, 2]>>> remove_last (5, [3, 5, 2, 5, 11]) [3, 5, 2, 11] IIIII "*** YOUR CODE HERE ***" Illustrated here is a more complete doctest that shows good testing methodology. It is a little cumbersome as documentation, but you'll want to think about it for your projects. Test every condition that might come up. Then you won't be surprised when...