Given a reference to a binary tree t. I am trying to write a python function using recursion that will check if an integer called k is stored in a leaf node in the tree. if not stored it should return false but if it is I want it to return True. starting function: def in_leaf(t,k):

write a code to Find the minimum node of a BST. Use either recursive or non-recursive approach.

Computer Science Write a simple LISP function to count all leaves in a tree. You need to use a double recursion for this problem. Your sample should produce this: (1 ( 2 3) (4 5 (6)) 7 (8)) ----> 8. Hint: Make sure not to count empty list as a leaf

Answer the given question with a proper explanation and step-by-step solution. the popBack() function fails when the node is empty the insertStudent only looks until a certain point

Algorithm 7 : Sift an elementprocedure sift( g : element );Input: a labelled branching A (with value A0);an element g;Output: an updated labelled branching A such that <labels(A)> = <labels(A0), g>;

Consider the following function:int mystery(NodeInt32* node){int counter = 0; while (node != NULL) {counter++;node = node->next; } return counter;}(a)The mystery function is recursive.A. True B. False

Draw the Recursion Execution Sequence. void preorder (tree_ptr ptr) { if (ptr) { (visit node); preorder (ptr->left_child); preorder (ptr->right_child); } }

Assume there is a rooted tree A. Write a recursive program that returns both the number of nodes(N) and number of edges(E) in A.

BST - Removal Algorithm 1. Complete the code of the iterative BST removal method (as seen on slide 9 of the lecture). Complete the • search for the node, and • the removal (case 1, case 2 and case 3) • the cases illustrated here might help you. public boolean remove(int target) { // where the item will be found. // This nd needs to be removed. BSTNode nd - root; BSTNode parent - null; // parent of the nd to be removed. boolean is left_child - false; // how nd was reached from parent. // search node with nd.data == target, // remember the parent, and if left or right path if (nd == null) // target not found, no removal. return false; else { // found target in nd, three cases for removal. // case 1: nd is a leaf, has no children // case 2: nd has 1 child at the left // case 2: nd has 1 child at the right // case 3: nd has two 2 children: replace nd by predecessor or by successor } Submit your solution as a pdf-file (neat handwriting is okay).