Write the following functions and add them to the BST class that was explained i
ID: 3805460 • Letter: W
Question
Write the following functions and add them to the BST class that was explained in the class:
a) A function to count the number of nodes in a binary tree (15 points)
b) A function to count the number of leaves (15 points)
c) A function to find the height of the tree (15 points)
Test your functions and include your source code and screenshots of the testing results.
Here is the Code we went over in class....
//*********************** genBST.h **************************
// generic binary search tree
#include <queue>
#include <stack>
#ifndef BINARY_SEARCH_TREE
#define BINARY_SEARCH_TREE
using namespace std;
template<class T>
class Stack : public stack<T> {
public:
T pop() {
T tmp = top();
stack<T>::pop();
return tmp;
}
};
template<class T>
class Queue : public queue<T> {
public:
T dequeue() {
T tmp = front();
queue<T>::pop();
return tmp;
}
void enqueue(const T& el) {
push(el);
}
};
template<class T> class BST;
template<class T>
class BSTNode {
public:
BSTNode() {
left = right = 0;
}
BSTNode(const T& e, BSTNode<T> *l = 0, BSTNode<T> *r = 0) {
el = e; left = l; right = r;
}
T el;
BSTNode<T> *left, *right;
};
template<class T>
class BST {
public:
BST() {
root = 0;
}
~BST() {
clear();
}
void clear() {
clear(root);
root = 0;
}
bool isEmpty() const {
return root == 0;
}
void preorder() {
preorder(root);
}
void inorder() {
inorder(root);
}
void postorder() {
postorder(root);
}
void insert(const T&);
void recursiveInsert(const T& el) {
recursiveInsert(root,el);
}
T* search(const T& el) const {
return search(root,el);
}
T* recursiveSearch(const T& el) const {
return recursiveSearch(root,el);
}
void deleteByCopying(BSTNode<T>*&);
void findAndDeleteByCopying(const T&);
void deleteByMerging(BSTNode<T>*&);
void findAndDeleteByMerging(const T&);
void iterativePreorder();
void iterativeInorder();
void iterativePostorder();
void breadthFirst();
void MorrisPreorder();
void MorrisInorder();
void MorrisPostorder();
void balance(T*,int,int);
protected:
BSTNode<T>* root;
void clear(BSTNode<T>*);
void recursiveInsert(BSTNode<T>*&, const T&);
T* search(BSTNode<T>*, const T&) const;
T* recursiveSearch(BSTNode<T>*, const T&) const;
void preorder(BSTNode<T>*);
void inorder(BSTNode<T>*);
void postorder(BSTNode<T>*);
virtual void visit(BSTNode<T>* p) {
cout << p->el<< ' ';
}
};
template<class T>
void BST<T>::clear(BSTNode<T> *p) {
if (p != 0) {
clear(p->left);
clear(p->right);
delete p;
}
}
template<class T>
void BST<T>::insert(const T& el) {
BSTNode<T> *p = root, *prev = 0;
while (p != 0) { // find a place for inserting new node;
prev = p;
if (el < p->el)
p = p->left;
else p = p->right;
}
if (root == 0) // tree is empty;
root = new BSTNode<T>(el);
else if (el < prev->el)
prev->left = new BSTNode<T>(el);
else prev->right = new BSTNode<T>(el);
}
template<class T>
void BST<T>::recursiveInsert(BSTNode<T>*& p, const T& el) {
if (p == 0)
p = new BSTNode<T>(el);
else if (el < p->el)
recursiveInsert(p->left, el);
else recursiveInsert(p->right,el);
}
template<class T>
T* BST<T>::search(BSTNode<T>* p, const T& el) const {
while (p != 0)
if (el == p->el)
return &p->el;
else if (el < p->el)
p = p->left;
else p = p->right;
return 0;
}
template<class T>
T* BST<T>::recursiveSearch(BSTNode<T>* p, const T& el) const {
if (p != 0)
if (el == p->el)
return &p->el;
else if (el < p->el)
return recursiveSearch(p->left,el);
else return recursiveSearch(p->right,el);
else return 0;
}
template<class T>
void BST<T>::inorder(BSTNode<T> *p) {
if (p != 0) {
inorder(p->left);
visit(p);
inorder(p->right);
}
}
template<class T>
void BST<T>::preorder(BSTNode<T> *p) {
if (p != 0) {
visit(p);
preorder(p->left);
preorder(p->right);
}
}
template<class T>
void BST<T>::postorder(BSTNode<T>* p) {
if (p != 0) {
postorder(p->left);
postorder(p->right);
visit(p);
}
}
template<class T>
void BST<T>::deleteByCopying(BSTNode<T>*& node) {
BSTNode<T> *previous, *tmp = node;
if (node->right == 0) // node has no right child;
node = node->left;
else if (node->left == 0) // node has no left child;
node = node->right;
else {
tmp = node->left; // node has both children;
previous = node; // 1.
while (tmp->right != 0) { // 2.
previous = tmp;
tmp = tmp->right;
}
node->el = tmp->el; // 3.
if (previous == node)
previous->left = tmp->left;
else previous->right = tmp->left; // 4.
}
delete tmp; // 5.
}
// findAndDeleteByCopying() searches the tree to locate the node containing
// el. If the node is located, the function DeleteByCopying() is called.
template<class T>
void BST<T>::findAndDeleteByCopying(const T& el) {
BSTNode<T> *p = root, *prev = 0;
while (p != 0 && !(p->el == el)) {
prev = p;
if (el < p->el)
p = p->left;
else p = p->right;
}
if (p != 0 && p->el == el)
if (p == root)
deleteByCopying(root);
else if (prev->left == p)
deleteByCopying(prev->left);
else deleteByCopying(prev->right);
else if (root != 0)
cout<< "el " << el << " is not in the tree ";
else cout << "the tree is empty ";
}
template<class T>
void BST<T>::deleteByMerging(BSTNode<T>*& node) {
BSTNode<T> *tmp = node;
if (node != 0) {
if (!node->right) // node has no right child: its left
node = node->left; // child (if any) is attached to its parent;
else if (node->left == 0) // node has no left child: its right
node = node->right; // child is attached to its parent;
else { // be ready for merging subtrees;
tmp = node->left; // 1. move left
while (tmp->right != 0)// 2. and then right as far as possible;
tmp = tmp->right;
tmp->right = // 3. establish the link between the
node->right; // the rightmost node of the left
// subtree and the right subtree;
tmp = node; // 4.
node = node->left; // 5.
}
delete tmp; // 6.
}
}
template<class T>
void BST<T>::findAndDeleteByMerging(const T& el) {
BSTNode<T> *node = root, *prev = 0;
while (node != 0) {
if (node->el == el)
break;
prev = node;
if (el < node->el)
node = node->left;
else node = node->right;
}
if (node != 0 && node->el == el)
if (node == root)
deleteByMerging(root);
else if (prev->left == node)
deleteByMerging(prev->left);
else deleteByMerging(prev->right);
else if (root != 0)
cout << "el " << el << " is not in the tree ";
else cout << "the tree is empty ";
}
template<class T>
void BST<T>::iterativePreorder() {
Stack<BSTNode<T>*> travStack;
BSTNode<T> *p = root;
if (p != 0) {
travStack.push(p);
while (!travStack.empty()) {
p = travStack.pop();
visit(p);
if (p->right != 0)
travStack.push(p->right);
if (p->left != 0) // left child pushed after right
travStack.push(p->left); // to be on the top of the stack;
}
}
}
template<class T>
void BST<T>::iterativeInorder() {
Stack<BSTNode<T>*> travStack;
BSTNode<T> *p = root;
while (p != 0) {
while (p != 0) { // stack the right child (if any)
if (p->right) // and the node itself when going
travStack.push(p->right); // to the left;
travStack.push(p);
p = p->left;
}
p = travStack.pop(); // pop a node with no left child
while (!travStack.empty() && p->right == 0) { // visit it and all nodes
visit(p); // with no right child;
p = travStack.pop();
}
visit(p); // visit also the first node with
if (!travStack.empty()) // a right child (if any);
p = travStack.pop();
else p = 0;
}
}
template<class T>
void BST<T>::iterativePostorder() {
Stack<BSTNode<T>*> travStack;
BSTNode<T>* p = root, *q = root;
while (p != 0) {
for ( ; p->left != 0; p = p->left)
travStack.push(p);
while (p->right == 0 || p->right == q) {
visit(p);
q = p;
if (travStack.empty())
return;
p = travStack.pop();
}
travStack.push(p);
p = p->right;
}
}
template<class T>
void BST<T>::breadthFirst() {
Queue<BSTNode<T>*> queue;
BSTNode<T> *p = root;
if (p != 0) {
queue.enqueue(p);
while (!queue.empty()) {
p = queue.dequeue();
visit(p);
if (p->left != 0)
queue.enqueue(p->left);
if (p->right != 0)
queue.enqueue(p->right);
}
}
}
template<class T>
void BST<T>::MorrisInorder() {
BSTNode<T> *p = root, *tmp;
while (p != 0)
if (p->left == 0) {
visit(p);
p = p->right;
}
else {
tmp = p->left;
while (tmp->right != 0 &&// go to the rightmost node of
tmp->right != p) // the left subtree or
tmp = tmp->right; // to the temporary parent of p;
if (tmp->right == 0) { // if 'true' rightmost node was
tmp->right = p; // reached, make it a temporary
p = p->left; // parent of the current root,
}
else { // else a temporary parent has been
visit(p); // found; visit node p and then cut
tmp->right = 0; // the right pointer of the current
p = p->right; // parent, whereby it ceases to be
} // a parent;
}
}
template<class T>
void BST<T>::MorrisPreorder() {
BSTNode<T> *p = root, *tmp;
while (p != 0) {
if (p->left == 0) {
visit(p);
p = p->right;
}
else {
tmp = p->left;
while (tmp->right != 0 &&// go to the rightmost node of
tmp->right != p) // the left subtree or
tmp = tmp->right; // to the temporary parent of p;
if (tmp->right == 0) { // if 'true' rightmost node was
visit(p); // reached, visit the root and
tmp->right = p; // make the rightmost node a temporary
p = p->left; // parent of the current root,
}
else { // else a temporary parent has been
tmp->right = 0; // found; cut the right pointer of
p = p->right; // the current parent, whereby it ceases
} // to be a parent;
}
}
}
template<class T>
void BST<T>::MorrisPostorder() {
BSTNode<T> *p = new BSTNode<T>(), *tmp, *q, *r, *s;
p->left = root;
while (p != 0)
if (p->left == 0)
p = p->right;
else {
tmp = p->left;
while (tmp->right != 0 &&// go to the rightmost node of
tmp->right != p) // the left subtree or
tmp = tmp->right; // to the temporary parent of p;
if (tmp->right == 0) { // if 'true' rightmost node was
tmp->right = p; // reached, make it a temporary
p = p->left; // parent of the current root,
}
else { // else a temporary parent has been found;
// process nodes between p->left (included) and p (excluded)
// extended to the right in modified tree in reverse order;
// the first loop descends this chain of nodes and reverses
// right pointers; the second loop goes back, visits nodes,
// and reverses right pointers again to restore the pointers
// to their original setting;
for (q = p->left, r = q->right, s = r->right;
r != p; q = r, r = s, s = s->right)
r->right = q;
for (s = q->right; q != p->left;
q->right = r, r = q, q = s, s = s->right)
visit(q);
visit(p->left); // visit node p->left and then cut
tmp->right = 0; // the right pointer of the current
p = p->right; // parent, whereby it ceases to be
} // a parent;
}
}
template<class T>
void BST<T>::balance (T data[], int first, int last) {
if (first <= last) {
int middle = (first + last)/2;
insert(data[middle]);
balance(data,first,middle-1);
balance(data,middle+1,last);
}
}
#endif
Explanation / Answer
Following is the implementation for AVL Tree Insertion. The following implementation uses the recursive BST insert to insert a new node. In the recursive BST insert, after insertion, we get pointers to all ancestors one by one in bottom up manner. So we don’t need parent pointer to travel up. The recursive code itself travels up and visits all the ancestors of the newly inserted node.
1) Perform the normal BST insertion.
2) The current node must be one of the ancestors of the newly inserted node. Update the height of the current node.
3) Get the balance factor (left subtree height – right subtree height) of the current node.
4) If balance factor is greater than 1, then the current node is unbalanced and we are either in Left Left case or left Right case. To check whether it is left left case or not, compare the newly inserted key with the key in left subtree root.
5) If balance factor is less than -1, then the current node is unbalanced and we are either in Right Right case or Right Left case. To check whether it is Right Right case or not, compare the newly inserted key with the key in right subtree root.
#include<stdio.h>
#include<stdlib.h>
// An AVL tree node
struct Node
{
int key;
struct Node *left;
struct Node *right;
int height;
};
// A utility function to get maximum of two integers
int max(int a, int b);
// A utility function to get height of the tree
int height(struct Node *N)
{
if (N == NULL)
return 0;
return N->height;
}
// A utility function to get maximum of two integers
int max(int a, int b)
{
return (a > b)? a : b;
}
/* Helper function that allocates a new node with the given key and
NULL left and right pointers. */
struct Node* newNode(int key)
{
struct Node* node = (struct Node*)
malloc(sizeof(struct Node));
node->key = key;
node->left = NULL;
node->right = NULL;
node->height = 1; // new node is initially added at leaf
return(node);
}
// A utility function to right rotate subtree rooted with y
// See the diagram given above.
struct Node *rightRotate(struct Node *y)
{
struct Node *x = y->left;
struct Node *T2 = x->right;
// Perform rotation
x->right = y;
y->left = T2;
// Update heights
y->height = max(height(y->left), height(y->right))+1;
x->height = max(height(x->left), height(x->right))+1;
// Return new root
return x;
}
// A utility function to left rotate subtree rooted with x
// See the diagram given above.
struct Node *leftRotate(struct Node *x)
{
struct Node *y = x->right;
struct Node *T2 = y->left;
// Perform rotation
y->left = x;
x->right = T2;
// Update heights
x->height = max(height(x->left), height(x->right))+1;
y->height = max(height(y->left), height(y->right))+1;
// Return new root
return y;
}
// Get Balance factor of node N
int getBalance(struct Node *N)
{
if (N == NULL)
return 0;
return height(N->left) - height(N->right);
}
// Recursive function to insert key in subtree rooted
// with node and returns new root of subtree.
struct Node* insert(struct Node* node, int key)
{
/* 1. Perform the normal BST insertion */
if (node == NULL)
return(newNode(key));
if (key < node->key)
node->left = insert(node->left, key);
else if (key > node->key)
node->right = insert(node->right, key);
else // Equal keys are not allowed in BST
return node;
/* 2. Update height of this ancestor node */
node->height = 1 + max(height(node->left),
height(node->right));
/* 3. Get the balance factor of this ancestor
node to check whether this node became
unbalanced */
int balance = getBalance(node);
// If this node becomes unbalanced, then
// there are 4 cases
// Left Left Case
if (balance > 1 && key < node->left->key)
return rightRotate(node);
// Right Right Case
if (balance < -1 && key > node->right->key)
return leftRotate(node);
// Left Right Case
if (balance > 1 && key > node->left->key)
{
node->left = leftRotate(node->left);
return rightRotate(node);
}
// Right Left Case
if (balance < -1 && key < node->right->key)
{
node->right = rightRotate(node->right);
return leftRotate(node);
}
/* return the (unchanged) node pointer */
return node;
}
// A utility function to print preorder traversal
// of the tree.
// The function also prints height of every node
void preOrder(struct Node *root)
{
if(root != NULL)
{
printf("%d ", root->key);
preOrder(root->left);
preOrder(root->right);
}
}
/* Drier program to test above function*/
int main()
{
struct Node *root = NULL;
/* Constructing tree given in the above figure */
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 30);
root = insert(root, 40);
root = insert(root, 50);
root = insert(root, 25);
/* The constructed AVL Tree would be
30
/
20 40
/
10 25 50
*/
printf("Preorder traversal of the constructed AVL"
" tree is ");
preOrder(root);
return 0;
}
Following is the implementation for AVL Tree Insertion. The following implementation uses the recursive BST insert to insert a new node. In the recursive BST insert, after insertion, we get pointers to all ancestors one by one in bottom up manner. So we don’t need parent pointer to travel up. The recursive code itself travels up and visits all the ancestors of the newly inserted node.
1) Perform the normal BST insertion.
2) The current node must be one of the ancestors of the newly inserted node. Update the height of the current node.
3) Get the balance factor (left subtree height – right subtree height) of the current node.
4) If balance factor is greater than 1, then the current node is unbalanced and we are either in Left Left case or left Right case. To check whether it is left left case or not, compare the newly inserted key with the key in left subtree root.
5) If balance factor is less than -1, then the current node is unbalanced and we are either in Right Right case or Right Left case. To check whether it is Right Right case or not, compare the newly inserted key with the key in right subtree root.
- C
- Java
#include<stdio.h>
#include<stdlib.h>
// An AVL tree node
struct Node
{
int key;
struct Node *left;
struct Node *right;
int height;
};
// A utility function to get maximum of two integers
int max(int a, int b);
// A utility function to get height of the tree
int height(struct Node *N)
{
if (N == NULL)
return 0;
return N->height;
}
// A utility function to get maximum of two integers
int max(int a, int b)
{
return (a > b)? a : b;
}
/* Helper function that allocates a new node with the given key and
NULL left and right pointers. */
struct Node* newNode(int key)
{
struct Node* node = (struct Node*)
malloc(sizeof(struct Node));
node->key = key;
node->left = NULL;
node->right = NULL;
node->height = 1; // new node is initially added at leaf
return(node);
}
// A utility function to right rotate subtree rooted with y
// See the diagram given above.
struct Node *rightRotate(struct Node *y)
{
struct Node *x = y->left;
struct Node *T2 = x->right;
// Perform rotation
x->right = y;
y->left = T2;
// Update heights
y->height = max(height(y->left), height(y->right))+1;
x->height = max(height(x->left), height(x->right))+1;
// Return new root
return x;
}
// A utility function to left rotate subtree rooted with x
// See the diagram given above.
struct Node *leftRotate(struct Node *x)
{
struct Node *y = x->right;
struct Node *T2 = y->left;
// Perform rotation
y->left = x;
x->right = T2;
// Update heights
x->height = max(height(x->left), height(x->right))+1;
y->height = max(height(y->left), height(y->right))+1;
// Return new root
return y;
}
// Get Balance factor of node N
int getBalance(struct Node *N)
{
if (N == NULL)
return 0;
return height(N->left) - height(N->right);
}
// Recursive function to insert key in subtree rooted
// with node and returns new root of subtree.
struct Node* insert(struct Node* node, int key)
{
/* 1. Perform the normal BST insertion */
if (node == NULL)
return(newNode(key));
if (key < node->key)
node->left = insert(node->left, key);
else if (key > node->key)
node->right = insert(node->right, key);
else // Equal keys are not allowed in BST
return node;
/* 2. Update height of this ancestor node */
node->height = 1 + max(height(node->left),
height(node->right));
/* 3. Get the balance factor of this ancestor
node to check whether this node became
unbalanced */
int balance = getBalance(node);
// If this node becomes unbalanced, then
// there are 4 cases
// Left Left Case
if (balance > 1 && key < node->left->key)
return rightRotate(node);
// Right Right Case
if (balance < -1 && key > node->right->key)
return leftRotate(node);
// Left Right Case
if (balance > 1 && key > node->left->key)
{
node->left = leftRotate(node->left);
return rightRotate(node);
}
// Right Left Case
if (balance < -1 && key < node->right->key)
{
node->right = rightRotate(node->right);
return leftRotate(node);
}
/* return the (unchanged) node pointer */
return node;
}
// A utility function to print preorder traversal
// of the tree.
// The function also prints height of every node
void preOrder(struct Node *root)
{
if(root != NULL)
{
printf("%d ", root->key);
preOrder(root->left);
preOrder(root->right);
}
}
/* Drier program to test above function*/
int main()
{
struct Node *root = NULL;
/* Constructing tree given in the above figure */
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 30);
root = insert(root, 40);
root = insert(root, 50);
root = insert(root, 25);
/* The constructed AVL Tree would be
30
/
20 40
/
10 25 50
*/
printf("Preorder traversal of the constructed AVL"
" tree is ");
preOrder(root);
return 0;
}
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