You have to write a code which will help to determine whether a binary tree is a max heap. The following code is a sample implementation of a binary tree, you can use it as a sample example. Code should print whether the tree is a max heap or not.
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I need help in being able to detect a max heap
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- in java : you need to start implementing a class for creating and storing Binary SearchTrees (BST). Each node of this BST will store the roll number, name and CGPA of a student.The class definitions will look like:class StudentBST;class StudentNode {friend class StudentBST;private:int rollNo;string name;Student’s CGPAStudentNode* left; // Pointer to the left subtree of a nodeStudentNode* right; // Pointer to the right subtree of a node};class StudentBST {private:StudentNode* root; // Pointer to the root node of the treepublic:StudentBST(); // Default constructor}; Write a menu-based driver function to illustrate the working of different functions of theStudentBST class. The menu should look like:1. Insert a new student2. Search for a student3. See the list of students4. QuitEnter your choice:in C++, create a binary search tree class using the class templatr below. class BinarySearchTree{ public:BinarySearchTree(); ~BinarySearchTree(); const int & findMin() const; const int & findMax() const; bool contains(cons tint &x) const; bool isEmpty(); void insert(const int &x); void remove(const int &x); private: struct BinaryNode{ int element; BinaryNode *left; BinaryNode *right; BinaryNode(const int &el, BinaryNode *lt, BinaryNode *rt) :element(el), left(lt), right(rt){} }; BinaryNode *root; void insert(const int &x, BinaryNode *&t) const; void remove(const int &x, BinaryNode *&t) const; BinaryNode * findMin(BinaryNode *t) const; BinaryNode * findMax(BinaryNode *t) const; bool contains(cons tint &x, BinaryNode *t) const; };The following class definitions to implement the Binary Search Tree is given. class Node ( public: int data; Node* left; Node* right; Node (int key) { data = key; left = NULL; right = NULL; }; class BST{ public: Node* root; BST () { root = NULL; } Node* insert (Node* root, int key); Node* search (Node* root, int key); Node* del (Node* root, int key); Node* findMin (Node* root); Node* findMax (Node* root); void inorder (Node* root); }; A. Write a program named as ProblemA_.cpp that implements the following functions using recursion. Make a menu-based implementation so that you can call each function based on the user input. insert: insert a new node in the BST Node* insert (Node* root, int key); print: prints the values of all items in the tree, using inorder traversal technique void inorder (Node* root); • search: search a node with a value in the BST Node* insert (Node* root, int key); findMinimum: returns the element with the minimum value in the BST Node* findMinimum (Node* root);…
- You are going to start implementing a classfor creating and storing Binary Search Trees(BST). Each node of this BST will store the roll number, name and CGPA of a student. The classdefinitions will look like: class StudentBST;class StudentNode {friend class StudentBST;private:int rollNo; // Student’s roll number (must be unique)string name; // Student’s namedouble cgpa; // Student’s CGPAStudentNode *left; // Pointer to the left subtree of a nodeStudentNode *right; // Pointer to the right subtree of a node};class StudentBST {private: 3 StudentNode *root; // Pointer to the root node of the BSTpublic:StudentBST(); // Default constructor};Implement the following two public member functions of the StudentBST class: bool insert (int rn, string n, double c)This function will insert a new student’s record in the StudentBST. The 3 arguments of thisfunction are the roll number, name, and CGPA of this new student, respectively. The insertioninto the tree will be done based upon the roll number of…Book{ AvlNode{ AvlTree{String title Book insert()int ID int height printTree()String author AvlNode left height()double price AvlNode right max()} } }From your understanding of AVL trees, implement the four different types of rotations and node swoping inside AvlTree.java; each question must be completed with inline Javadoc and/or comments:1. rotRight(AvlNode someNode) and returns AvlNode.2. rotLeft(AvlNode someNode) and returns AvlNode.3. doubleRotLeft(AvlNode someNode) and returns AvlNode. 4. doubleRotRight(AvlNode someNode) and returns AvlNode. 5. Complete a driver code under the main methodAssignment 4 The goal of this assignment is to write a Java program that finds the lowest common ancestor of two nodes in a binary tree. To accomplish this goal, you will have to implement a program that allows the user to ask for two letters on the tree shown below, and finds the lowest common ancestor of those two nodes in the tree.
- Write a recursive function, OnlyChild(..), that returns the number of nodes in a binary tree that has only one child. Consider binaryTrecNode structure is defined as the following. struct binaryTreeNode int info; binaryTreeNode *llink; binaryTreeNode *rlink; The function is declared as the following. You ust write the function as a recursive function. You will not get any credits if a non-recursive solution is used. int OnlyChild(binaryTreeNode *p); For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). B IUS Paragraph Arial 10pt 11Computer Science JAVA Write a program that maintains the names of your friends and relatives and thus serves as a friends list. You should be able to enter, delete, modify, or search this data. You should assume that the names are unique. use a class to represent the names in the friends list and another class to represent the friends list itself. This class should contain a Binary Search Tree of names as a data field. (TreeNode Class BinarySearchTree Class FriendsList Class)A tree is implemented using a node structure defined as: struct node{ int data; struct node *left; strict node *right; }; Write a function whose prototype is: int smallest(struct node *); which accepts a tree (pointer to the root) and returns the smallest node in the right sub-tree of the root. If root is NULL, or if there is no right sub-tree, returm the number -999.
- In c++, Write a Binary Search Tree for strings datatype, using pointers Besure that it has a method that can Rebalance the tree. See below for driver file Testing requirments: int main() { myBTS BTS1; // Declare/Instanitate an instance of the BTS BTS1.inTree("Fred"); BTS1.inTree("Able"); BTS1.inTree("Tuyet"); BTS1.inTree("Mojo"); BTS1.inTree("Linda"); BTS1.inTree("Leena"); BTS1.inTree("Xoe"); BTS1.inTree("Zohe"); BTS1.inTree("Alfred"); BTS1.inTree("Thanos"); BTS1.inTree("koji"); BTS1.inTree("Hally"); BTS1.inTree("Lee"); BTS1.inTree("Mode"); cout << "Number of names in the tree: " << BTS1.count() << endl; if ( BTS1.verify () ) // Function returns a Boolean value { cout << "BST Verified" <please convert to C languange #include<bits/stdc++.h>using namespace std; class tree{ //tree node public: int data; tree *left; tree *right;}; bool hasRootToLeafSum(tree *root, int s){ bool path=false; //declare boolean variable path //base condition checking if(root==NULL && s==0) return true; s-=root->data; //subtract current root value //checking whether leaf node reached and remaining sum =0 if(s==0 && root->left==NULL && root->right==NULL) return true; //recursively done for both subtrees if(root->left){//for left subtree path=path||hasRootToLeafSum(root->left, s); } if(root->right){//for right subtree path=path||hasRootToLeafSum(root->right, s); } return path;} tree* newnode(int data){ //creating new nodes tree* node = (tree*)malloc(sizeof(tree)); node->data = data; node->left = NULL; node->right = NULL;…Language C++: Kill Monger Suppose you are playing a killing game and have to kill few of your enemies. You are at acliff and from there you can see few of them. You see the enemies at your height only.Therefore, you can only kill at the same height. The enemies are forming a balanced binarytree where their General is at the top of the other cliff, his two commanders are on the firstlevel below him. The soldiers are at the lowest level. Total four soldiers are appointed. Underboth the commanders exact two soldiers serve.You have to make a strategy to kill the left commander first, than whoever takes his positionkill him. Now start killing the right side with same strategy until only one enemy is left at leftside. Now increase your height and kill the General. You can only win if only three of yourenemies are left.Display the enemies who are left and who have been killed. Don’t forget to print your victorymessage.HINT: Balancing is the key to achieve your goal.