Question

For AVL tree, write C functions to rotate-right, rotate-left, insert and delete, to
check if a BST is a AVL tree.

Answer 1

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;
}
  
  
  
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;
}

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;
}

int getBalance(Node *N)
{
if (N == NULL)
return 0;
return height(N->left) –
height(N->right);
}

Node* deleteNode(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
{
// node with only one child or no child
if( (root->left == NULL) ||
(root->right == NULL) )
{
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)
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 = 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;
}

/* Returns true if binary tree with root as root is height-balanced which means AVL tree*/
bool isBalanced(struct node *root)
{
int lh; /* for height of left subtree */
int rh; /* for height of right subtree */
  
/* If tree is empty then return true */
if(root == NULL)
return 1;
  
/* Get the height of left and right sub trees */
lh = height(root->left);
rh = height(root->right);
  
if( abs(lh-rh) <= 1 &&
isBalanced(root->left) &&
isBalanced(root->right))
return 1;
  
/* If we reach here then tree is not height-balanced */
return 0;
}
  
/* UTILITY FUNCTIONS TO TEST isBalanced() FUNCTION */
  
/* returns maximum of two integers */
int max(int a, int b)
{
return (a >= b)? a: b;
}