Solve T(n) = 4T(n/2) + O(n³) using the recursion tree method. • number of subproblems/nodes at depth d: workload per subproblem/node at depth d:
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- The BST remove algorithm traverses the tree from the root to find the node to remove. When the node being removed has 2 children, the node's successor is found and a recursive call is made. One node is visited per level, and in the worst-case scenario, the tree is traversed twice from the root to a leaf. A BST with N nodes has at least log2N levels and at most N levels. Therefore, the runtime complexity of removal is best case O(logN) and worst case O(N). Two pointers are used to traverse the tree during removal. When the node being removed has 2 children, a third pointer and a copy of one node's data are also used, and one recursive call is made. Thus, the space complexity of removal is always O(1)." I have to explain this clearly! and the advantages of the BST algorithimFor each, draw the recursion tree, find the height of the tree, the running time of each layer, and the sum of running times. Then use this info to find the explicit answer for T(n). a. T(n) = 2T(n/4) + √ n (n is a power of 4 (n = 4^k) for some positive integer k) b. T(n) = 9T(n/3) + n^2 (n is a power of 3 (n = 3^k) for some positive integer k) c. T(n) = T(n/2) + 1 (n is a power of 2 (n = 2^k) for some positive integer k)Provide an example of a recursive function in which the amount of work on each activation is constant. Provide the recurrence equation and the initial condition that counts the number of operations executed. Specify which operations you are counting and why they are the critical ones to count to assess its execution time. Draw the recursion tree for that function and determine the Big-Θ by determining a formula that counts the number of nodes in the tree as a function of n.
- Create a recursion tree for T(n)=3T(n/2)+2n^2, with T(1)=c2, for n=8 . calculates the total number of comparisons.Using your knowledge in binary tree traversals, implement the following tree traversing algorithms in either Java or C++, i. inorder, ii. preorder iii. and postorder traversalGiven the following non-recursive implementation of depth-first search: A. Complete the implementation of depth-first search by filling in the TODO sections with the appropriate C++ code. Remember to: Print out each node you visit. Visit each node exactly once. B. What is the purpose of stack<string> q? C. What is the purpose of set<string> v?
- Given a binary tree with integer data at all nodes (including leaves), your task is to create a pretty printer. Each node in the tree can have 0, 1, or 2 children in python3. constructor that takes a required int argument for the data stored in the node, and two optional Node arguments for the left and right children, respectively. __str__(self) -> str: returns a string representation of the subtree rooted at the self node [must use recursion] compute_height(self) -> int: returns the height of the current node. The height is the distance of the longest path between a leaf node and the current node. [must use recursion] pretty_print(self, indent: int): prints a formatted tree string with each node on a separate line, using the format value: height, and making sure that sibling and cousin nodes (i.e., those at the same distance from the root node) are indented using the same number of spaces (see example below). The root is indented 0 spaces, the children of the root are indented…Consider the following algorithm: int f(n) /* n is a positive integer */ if (n <=3) return n int sum = f(n-1) if (n is even) return sum + f(n-2) else return sum + f(n-3) Trace execution of f(6) by drawing the recursion tree. Show all function calls (callers and called functions), and intermediate results; and show what f(6) returns. Make sure that you draw all subtrees even if some are identical.Use a simple recursive technique to determine whether a binary tree is BST?
- In java and in O(logn) time Write a method that balances an existing BST, call it balance(). A BST is balanced if the height of its left and right sub-trees are different by at most one. Recursively applied. If the tree is balanced, then searching for keys will take act like binary search and require only logn comparisons. No performance requirements on your balancing algorithm. (Come up with a way yourself - don't skip to 3.3. That section is really complicated and meant for the harder case where you need to do it in log time.)Write a recursive programme that accepts a BST and an integer k as input and returns the tree's kth smallest node. Remember that the nodes in the left subtree are all smaller than the root, while those in the right are all bigger.A binomial tree, Bn is defined recursively as follows. B0 is the tree with a single vertex.Create Bn+1, where n is a nonnegative integer, by making two copies of Bn; the first copy becomes the root tree of Bn+1, and the second copy becomes the leftmost child of the root in the first copy.Here are examples for n = 0 to 3: A. Create a table that has the number of nodes in each depth, d, of B0 to B4, where d ≥ 1 (you should NOT have to draw B5!). B. What do you think the answers for problem d, above, for B5?