Is there a smaller-caller in this method? Does the recursive call to the algorithm involve a smaller case of the original problem? private int recSize(BSTNode<T> node) // Returns the number of elements in subtree rooted at node. { if (node == null)      return 0; else    return 1 + recSize(node.getLeft()) + recSize(node.getRight()); }

bool DelOddCopEven(Node* headPtr) {    if (headPtr == nullptr) {        return false;    }     Node* prev = nullptr;    Node* current = headPtr;     while (current != nullptr) {        if (current->data % 2 != 0) {            // Odd-valued node, delete it            if (prev != nullptr) {                prev->link = current->link;            } else {                headPtr = current->link;            }            Node* temp = current;            current = current->link;            delete temp;        } else {            // Even-valued node, create a copy and insert it after the original node            Node* newNode = new Node;            newNode->data = current->data;            newNode->link = current->link;            current->link = newNode;             // Update the previous and current pointers            prev = newNode;            current = newNode->link;        }    }     return true;} int FindListLength(Node* headPtr){   int length = 0;…

Write a code to implement binary search tree. Implement all the methods, find, insert, delete, minimum, max, find successor. Test out all the methods.

Write a method isBST() that takes a Node as argument and returns true if the argument node is the root of a binary search tree, false otherwise.Hint : This task is also more difficult than it might seem, because the order in which youcall the methods in the previous three exercises is important