AVL Tree : You are required to develop an AVL tree ( class AvlNode) to support t
ID: 3724587 • Letter: A
Question
AVL Tree
:
You are required to develop an AVL tree (
class AvlNode)
to support the following
operations
Find method
:
returns the node that contains the value indicated by the input
parameter;
Insert method:
insert the specified value into the AVL tree;
Remove method: remove the specified value from the AVL tree;
Inorder traversal method.
Testing data
You may test your program by do the following:
Insert 3, 2, 1, 4, 5, 6, 7, 16,
15, 14, 13, 12, 11, 10, 8, 9 into an initially empty AVL
tree, then
print the tree using Inorder traversal;
Remove 9, 6, 1, 3, then print the tree using Inorder traversal.
Explanation / Answer
// C program to delete a node from AVL Tree
#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
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);
}
//function to find a node
struct Node* findNode(struct Node* root, int key){
if(root == NULL){
return root;
}
// If the key to be deleted is smaller than the
// root's key, then it lies in left subtree
if ( key < root->key )
root->left = findNode(root->left, key);
// If the key to be deleted is greater than the
// root's key, then it lies in right subtree
else if( key > root->key )
root->right = findNode(root->right, key);
else return root;
}
struct Node* find(struct Node* node, int element){
if(node != NULL){
if(node->key == element){
return node;
} else {
struct Node* foundNode = find(node->left, element);
if(foundNode == NULL) {
foundNode = find(node->right, element);
}
return foundNode;
}
} else {
return NULL;
}
}
struct Node* insert(struct Node* node, int key)
{
/* 1. Perform the normal BST rotation */
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 not allowed
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;
}
/* Given a non-empty binary search tree, return the
node with minimum key value found in that tree.
Note that the entire tree does not need to be
searched. */
struct Node * minValueNode(struct Node* node)
{
struct Node* current = node;
/* loop down to find the leftmost leaf */
while (current->left != NULL)
current = current->left;
return current;
}
// Recursive function to delete a node with given key
// from subtree with given root. It returns root of
// the modified subtree.
struct Node* removeNode(struct Node* root, int key)
{
// STEP 1: PERFORM STANDARD BST DELETE
if (root == NULL)
return root;
// If the key to be deleted is smaller than the
// root's key, then it lies in left subtree
if ( key < root->key )
root->left = removeNode(root->left, key);
// If the key to be deleted is greater than the
// root's key, then it lies in right subtree
else if( key > root->key )
root->right = removeNode(root->right, key);
// if key is same as root's key, then This is
// the node to be deleted
else
{
// node with only one child or no child
if( (root->left == NULL) || (root->right == NULL) )
{
struct Node *temp = root->left ? root->left :
root->right;
// No child case
if (temp == NULL)
{
temp = root;
root = NULL;
}
else // One child case
*root = *temp; // Copy the contents of
// the non-empty child
free(temp);
}
else
{
// node with two children: Get the inorder
// successor (smallest in the right subtree)
struct Node* temp = minValueNode(root->right);
// Copy the inorder successor's data to this node
root->key = temp->key;
// Delete the inorder successor
root->right = removeNode(root->right, temp->key);
}
}
// If the tree had only one node then return
if (root == NULL)
return root;
// STEP 2: UPDATE HEIGHT OF THE CURRENT NODE
root->height = 1 + max(height(root->left),
height(root->right));
// STEP 3: GET THE BALANCE FACTOR OF THIS NODE (to
// check whether this node became unbalanced)
int balance = getBalance(root);
// If this node becomes unbalanced, then there are 4 cases
// Left Left Case
if (balance > 1 && getBalance(root->left) >= 0)
return rightRotate(root);
// Left Right Case
if (balance > 1 && getBalance(root->left) < 0)
{
root->left = leftRotate(root->left);
return rightRotate(root);
}
// Right Right Case
if (balance < -1 && getBalance(root->right) <= 0)
return leftRotate(root);
// Right Left Case
if (balance < -1 && getBalance(root->right) > 0)
{
root->right = rightRotate(root->right);
return leftRotate(root);
}
return root;
}
/* Given a binary tree, print its nodes in inorder*/
void inorder(struct Node* node)
{
if (node == NULL)
return;
/* first recur on left child */
inorder(node->left);
/* then print the data of node */
printf("%d ", node->key);
/* now recur on right child */
inorder(node->right);
}
/* Drier program to test above function*/
int main()
{
struct Node *root = NULL;
struct Node *node = NULL;
/* Constructing tree */
root = insert(root, 3);
root = insert(root, 2);
root = insert(root, 1);
root = insert(root, 4);
root = insert(root, 5);
root = insert(root, 6);
root = insert(root, 7);
root = insert(root, 16);
root = insert(root, 15);
root = insert(root, 14);
root = insert(root, 13);
root = insert(root, 12);
root = insert(root, 11);
root = insert(root, 10);
root = insert(root, 8);
root = insert(root, 9);
printf("Inorder traversal of the constructed AVL "
"tree is ");
inorder(root);
root = removeNode(root, 9);
root = removeNode(root, 6);
root = removeNode(root, 1);
root = removeNode(root, 3);
printf(" Inorder traversal after deletion ");
inorder(root);
return 0;
}
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