AVL with duplicate keys

// C++ program of AVL tree that
// handles duplicates
#include
#include

// An AVL tree node
struct node {
int key;
struct node* left;
struct node* right;
int height;
int count;
};

// 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
node->count = 1;
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);
}

struct node* insert(struct node* node, int key)
{
/* 1. Perform the normal BST rotation */
if (node == NULL)
return (newNode(key));

// If key already exists in BST, increment count and return
if (key == node->key) {
(node->count)++;
return node;
}

/* Otherwise, recur down the tree */
if (key < node->key)
node->left = insert(node->left, key);
else
node->right = insert(node->right, key);

/* 2. Update height of this ancestor node */
node->height = max(height(node->left), height(node->right)) + 1;

/* 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;
}

struct node* deleteNode(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 = deleteNode(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 = deleteNode(root->right, key);

// if key is same as root’s key, then This is the node
// to be deleted
else {
// If key is present more than once, simply decrement
// count and return
if (root->count > 1) {
(root->count)–;
return;
}
// Else, delete the node

// 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 = deleteNode(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 = max(height(root->left), height(root->right)) + 1;

// 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;
}

// 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(%d) “, root->key, root->count);
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, 9);
root = insert(root, 5);
root = insert(root, 10);
root = insert(root, 5);
root = insert(root, 9);
root = insert(root, 7);
root = insert(root, 17);

printf(“Pre order traversal of the constructed AVL tree is \n”);
preOrder(root);

root = deleteNode(root, 9);

printf(“\nPre order traversal after deletion of 9 \n”);
preOrder(root);

return 0;
}