#include #include // 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); } 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 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 { // 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 ", 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, 9); root = insert(root, 5); root = insert(root, 10); root = insert(root, 0); root = insert(root, 6); root = insert(root, 11); root = insert(root, -1); root = insert(root, 1); root = insert(root, 2); /* The constructed AVL Tree would be 9 / \ 1 10 / \ \ 0 5 11 / / \ -1 2 6 */ printf("Pre order traversal of the constructed AVL tree is \n"); preOrder(root); root = deleteNode(root, 10); /* The AVL Tree after deletion of 10 1 / \ 0 9 / / \ -1 5 11 / \ 2 6 */ printf("\nPre order traversal after deletion of 10 \n"); preOrder(root); return 0; }