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AVL with duplicate keys
• Difficulty Level : Medium
• Last Updated : 07 Feb, 2020

This is to augment AVL tree node to store count together with regular fields like key, left and right pointers.
Insertion of keys 12, 10, 20, 9, 11, 10, 12, 12 in an empty Binary Search Tree would create following.

```          12(3)
/        \
10(2)      20(1)
/    \
9(1)   11(1)
```

Count of a key is shown in bracket

Below is implementation of normal AVL Tree with count with every key. This code basically is taken from code for insert and delete in AVL tree. The changes made for handling duplicates are highlighted, rest of the code is same.

The important thing to note is changes are very similar to simple Binary Search Tree changes.

## C

 `// 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 and update the count ` `            ``root->key = temp->key; ` `            ``root->count = temp->count; ` `            ``temp->count = 1; ` ` `  `            ``// 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); ` `    ``} ` `} ` ` `  `/* Driver 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; ` `} `

## Java

 `// Java program of AVL tree that handles duplicates ` `import` `java.util.*; ` ` `  `class` `solution { ` ` `  `    ``// An AVL tree node ` `    ``static` `class` `node { ` `        ``int` `key; ` `        ``node left; ` `        ``node right; ` `        ``int` `height; ` `        ``int` `count; ` `    ``} ` ` `  `    ``// A utility function to get height of the tree ` `    ``static` `int` `height(node N) ` `    ``{ ` `        ``if` `(N == ``null``) ` `            ``return` `0``; ` `        ``return` `N.height; ` `    ``} ` ` `  `    ``// A utility function to get maximum of two integers ` `    ``static` `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. */` `    ``static` `node newNode(``int` `key) ` `    ``{ ` `        ``node node = ``new` `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. ` `    ``static` `node rightRotate(node y) ` `    ``{ ` `        ``node x = y.left; ` `        ``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. ` `    ``static` `node leftRotate(node x) ` `    ``{ ` `        ``node y = x.right; ` `        ``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 ` `    ``static` `int` `getBalance(node N) ` `    ``{ ` `        ``if` `(N == ``null``) ` `            ``return` `0``; ` `        ``return` `height(N.left) - height(N.right); ` `    ``} ` ` `  `    ``static` `node insert(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. */` `    ``static` `node minValueNode(node node) ` `    ``{ ` `        ``node current = node; ` ` `  `        ``/* loop down to find the leftmost leaf */` `        ``while` `(current.left != ``null``) ` `            ``current = current.left; ` ` `  `        ``return` `current; ` `    ``} ` ` `  `    ``static` `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` `{ ` `            ``// If key is present more than once, simply decrement ` `            ``// count and return ` `            ``if` `(root.count > ``1``) { ` `                ``(root.count)--; ` `                ``return` `null``; ` `            ``} ` `            ``// ElSE, delete the node ` ` `  `            ``// node with only one child or no child ` `            ``if` `((root.left == ``null``) || (root.right == ``null``)) { ` `                ``node temp = root.left != ``null` `? 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 ` `            ``} ` `            ``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 and update the count ` `                ``root.key = temp.key; ` `                ``root.count = temp.count; ` `                ``temp.count = ``1``; ` ` `  `                ``// 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 ` `    ``static` `void` `preOrder(node root) ` `    ``{ ` `        ``if` `(root != ``null``) { ` `            ``System.out.printf(``"%d(%d) "``, root.key, root.count); ` `            ``preOrder(root.left); ` `            ``preOrder(root.right); ` `        ``} ` `    ``} ` ` `  `    ``/* Driver program to test above function*/` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``node root = ``null``; ` ` `  `        ``/* Coning 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``); ` ` `  `        ``System.out.printf(``"Pre order traversal of the constructed AVL tree is \n"``); ` `        ``preOrder(root); ` ` `  `        ``deleteNode(root, ``9``); ` ` `  `        ``System.out.printf(``"\nPre order traversal after deletion of 9 \n"``); ` `        ``preOrder(root); ` `    ``} ` `} ` `// contributed by Arnab Kundu `

## C#

 `// C# program of AVL tree that ` `// handles duplicates ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// An AVL tree node ` `    ``class` `node { ` `        ``public` `int` `key; ` `        ``public` `node left; ` `        ``public` `node right; ` `        ``public` `int` `height; ` `        ``public` `int` `count; ` `    ``} ` ` `  `    ``// A utility function to get ` `    ``// height of the tree ` `    ``static` `int` `height(node N) ` `    ``{ ` `        ``if` `(N == ``null``) ` `            ``return` `0; ` `        ``return` `N.height; ` `    ``} ` ` `  `    ``// A utility function to get ` `    ``// maximum of two integers ` `    ``static` `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. */` `    ``static` `node newNode(``int` `key) ` `    ``{ ` `        ``node node = ``new` `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. ` `    ``static` `node rightRotate(node y) ` `    ``{ ` `        ``node x = y.left; ` `        ``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. ` `    ``static` `node leftRotate(node x) ` `    ``{ ` `        ``node y = x.right; ` `        ``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 ` `    ``static` `int` `getBalance(node N) ` `    ``{ ` `        ``if` `(N == ``null``) ` `            ``return` `0; ` `        ``return` `height(N.left) - height(N.right); ` `    ``} ` ` `  `    ``static` `node insert(node node, ``int` `key) ` `    ``{ ` `        ``/*1. Perform the normal BST rotation */` `        ``if` `(node == ``null``) ` `            ``return` `(newNode(key)); ` ` `  `        ``// If key already exists in BST, ` `        ``// icnrement 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. */` `    ``static` `node minValueNode(node node) ` `    ``{ ` `        ``node current = node; ` ` `  `        ``/* loop down to find the ` `       ``leftmost leaf */` `        ``while` `(current.left != ``null``) ` `            ``current = current.left; ` ` `  `        ``return` `current; ` `    ``} ` ` `  `    ``static` `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` `{ ` `            ``// If key is present more than ` `            ``// once, simply decrement ` `            ``// count and return ` `            ``if` `(root.count > 1) { ` `                ``(root.count)--; ` `                ``return` `null``; ` `            ``} ` ` `  `            ``// ElSE, delete the node ` ` `  `            ``// node with only one child ` `            ``// or no child ` `            ``if` `((root.left == ``null``) || (root.right == ``null``)) { ` `                ``node temp = root.left != ``null` `? 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 ` `            ``} ` `            ``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 and update the count ` `                ``root.key = temp.key; ` `                ``root.count = temp.count; ` `                ``temp.count = 1; ` ` `  `                ``// 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 ` `    ``static` `void` `preOrder(node root) ` `    ``{ ` `        ``if` `(root != ``null``) { ` `            ``Console.Write(root.key + ``"("` `+ root.count + ``") "``); ` `            ``preOrder(root.left); ` `            ``preOrder(root.right); ` `        ``} ` `    ``} ` ` `  `    ``// Driver Code ` `    ``static` `public` `void` `Main(String[] args) ` `    ``{ ` `        ``node root = ``null``; ` ` `  `        ``/* Coning 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); ` ` `  `        ``Console.Write(``"Pre order traversal of "` `                      ``+ ``"the constructed AVL tree is \n"``); ` `        ``preOrder(root); ` ` `  `        ``deleteNode(root, 9); ` ` `  `        ``Console.Write(``"\nPre order traversal after "` `                      ``+ ``"deletion of 9 \n"``); ` `        ``preOrder(root); ` `    ``} ` `} ` ` `  `// This code is contributed by Arnab Kundu `

Output:

```Pre order traversal of the constructed AVL tree is
9(2) 5(2) 7(1) 10(1) 17(1)
Pre order traversal after deletion of 9
9(1) 5(2) 7(1) 10(1) 17(1) ```

Thanks to Rounaq Jhunjhunu Wala for sharing initial code. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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