AVL Tree | Set 2 (Deletion) - GeeksforGeeks

11 Apr, 2022
We have discussed AVL insertion in the previous post. In this post, we will follow a similar approach for deletion.
Steps to follow for deletion. To make sure that the given tree remains AVL after every deletion, we must augment the standard BST delete operation to perform some re-balancing. Following are two basic operations that can be performed to re-balance a BST without violating the BST property (keys(left) < key(root) < keys(right)). 1) Left Rotation 2) Right Rotation 
T1, T2 and T3 are subtrees of the tree rooted with y (on left side)
or x (on right side)
                y                               x
               / \     Right Rotation          /  \
              x   T3   – - – - – - – >        T1   y
             / \       < - - - - - - -            / \
            T1  T2     Left Rotation            T2  T3
Keys in both of the above trees follow the following order
      keys(T1) < key(x) < keys(T2) < key(y) < keys(T3)
So BST property is not violated anywhere.
Let w be the node to be deleted 1) Perform standard BST delete for w. 2) Starting from w, travel up and find the first unbalanced node. Let z be the first unbalanced node, y be the larger height child of z, and x be the larger height child of y. Note that the definitions of x and y are different from insertion here. 3) Re-balance the tree by performing appropriate rotations on the subtree rooted with z. There can be 4 possible cases that needs to be handled as x, y and z can be arranged in 4 ways. Following are the possible 4 arrangements: a) y is left child of z and x is left child of y (Left Left Case) b) y is left child of z and x is right child of y (Left Right Case) c) y is right child of z and x is right child of y (Right Right Case) d) y is right child of z and x is left child of y (Right Left Case)Like insertion, following are the operations to be performed in above mentioned 4 cases. Note that, unlike insertion, fixing the node z won’t fix the complete AVL tree. After fixing z, we may have to fix ancestors of z as well (See this video lecture for proof)
a) Left Left Case 
T1, T2, T3 and T4 are subtrees.
         z                                      y 
        / \                                   /   \
       y   T4      Right Rotate (z)          x      z
      / \          - - - - - - - - ->      /  \    /  \ 
     x   T3                               T1  T2  T3  T4
    / \
  T1   T2
b) Left Right Case 
     z                               z                           x
    / \                            /   \                        /  \ 
   y   T4  Left Rotate (y)        x    T4  Right Rotate(z)    y      z
  / \      - - - - - - - - ->    /  \      - - - - - - - ->  / \    / \
T1   x                          y    T3                    T1  T2 T3  T4
    / \                        / \
  T2   T3                    T1   T2
c) Right Right Case 
  z                                y
 /  \                            /   \ 
T1   y     Left Rotate(z)       z      x
    /  \   - - - - - - - ->    / \    / \
   T2   x                     T1  T2 T3  T4
       / \
     T3  T4
d) Right Left Case 
   z                            z                            x
  / \                          / \                          /  \ 
T1   y   Right Rotate (y)    T1   x      Left Rotate(z)   z      y
    / \  - - - - - - - - ->     /  \   - - - - - - - ->  / \    / \
   x   T4                      T2   y                  T1  T2  T3  T4
  / \                              /  \
T2   T3                           T3   T4
Unlike insertion, in deletion, after we perform a rotation at z, we may have to perform a rotation at ancestors of z. Thus, we must continue to trace the path until we reach the root.
Example:  
A node with value 32 is being deleted. After deleting 32, we travel up and find the first unbalanced node which is 44. We mark it as z, its higher height child as y which is 62, and y’s higher height child as x which could be either 78 or 50 as both are of same height. We have considered 78. Now the case is Right Right, so we perform left rotation.
C implementation Following is the C implementation for AVL Tree Deletion. The following C implementation uses the recursive BST delete as basis. In the recursive BST delete, after deletion, we get pointers to all ancestors one by one in bottom up manner. So we don’t need parent pointer to travel up. The recursive code itself travels up and visits all the ancestors of the deleted node. 1) Perform the normal BST deletion. 2) The current node must be one of the ancestors of the deleted node. Update the height of the current node. 3) Get the balance factor (left subtree height – right subtree height) of the current node. 4) If balance factor is greater than 1, then the current node is unbalanced and we are either in Left Left case or Left Right case. To check whether it is Left Left case or Left Right case, get the balance factor of left subtree. If balance factor of the left subtree is greater than or equal to 0, then it is Left Left case, else Left Right case. 5) If balance factor is less than -1, then the current node is unbalanced and we are either in Right Right case or Right Left case. To check whether it is Right Right case or Right Left case, get the balance factor of right subtree. If the balance factor of the right subtree is smaller than or equal to 0, then it is Right Right case, else Right Left case.
C++
C
Java
Python3
C#
Javascript
// C++ program to delete a node from AVL Tree#include<bits/stdc++.h>using namespace std; // An AVL tree nodeclass Node{    public:    int key;    Node *left;    Node *right;    int height;}; // A utility function to get maximum// of two integersint max(int a, int b); // A utility function to get height// of the treeint height(Node *N){    if (N == NULL)        return 0;    return N->height;} // A utility function to get maximum// of two integersint 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. */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    return(node);} // A utility function to right// rotate subtree rooted with y// See the diagram given above.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.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 Nint getBalance(Node *N){    if (N == NULL)        return 0;    return height(N->left) -           height(N->right);} Node* insert(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 valuefound in that tree. Note that the entiretree does not need to be searched. */Node * minValueNode(Node* node){    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.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;} // A utility function to print preorder// traversal of the tree.// The function also prints height// of every nodevoid preOrder(Node *root){    if(root != NULL)    {        cout << root->key << " ";        preOrder(root->left);        preOrder(root->right);    }} // Driver Codeint main(){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    */     cout << "Preorder 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    */     cout << "\nPreorder traversal after"         << " deletion of 10 \n";    preOrder(root);     return 0;} // This code is contributed by rathbhupendra
// C program to delete a node from AVL Tree#include<stdio.h>#include<stdlib.h> // An AVL tree nodestruct Node{    int key;    struct Node *left;    struct Node *right;    int height;}; // A utility function to get maximum of two integersint max(int a, int b); // A utility function to get height of the treeint height(struct Node *N){    if (N == NULL)        return 0;    return N->height;} // A utility function to get maximum of two integersint 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 Nint 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 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* 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 = 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;} // A utility function to print preorder traversal of// the tree.// The function also prints height of every nodevoid preOrder(struct Node *root){    if(root != NULL)    {        printf("%d ", root->key);        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, 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("Preorder 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("\nPreorder traversal after deletion of 10 \n");    preOrder(root);     return 0;}
// Java program for deletion in AVL Tree class Node{    int key, height;    Node left, right;     Node(int d)    {        key = d;        height = 1;    }} class AVLTree{    Node root;     // A utility function to get height of the tree    int height(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;    }     // A utility function to right rotate subtree rooted with y    // See the diagram given above.    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.    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    int getBalance(Node N)    {        if (N == null)            return 0;        return height(N.left) - height(N.right);    }     Node insert(Node node, int key)    {        /* 1. Perform the normal BST rotation */        if (node == null)            return (new Node(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        Wunbalanced */        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. */    Node minValueNode(Node node)    {        Node current = node;         /* loop down to find the leftmost leaf */        while (current.left != null)        current = current.left;         return current;    }     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 = null;                if (temp == root.left)                    temp = root.right;                else                    temp = root.left;                 // 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                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(Node node)    {        if (node != null)        {            System.out.print(node.key + " ");            preOrder(node.left);            preOrder(node.right);        }    }     public static void main(String[] args)    {        AVLTree tree = new AVLTree();         /* Constructing tree given in the above figure */        tree.root = tree.insert(tree.root, 9);        tree.root = tree.insert(tree.root, 5);        tree.root = tree.insert(tree.root, 10);        tree.root = tree.insert(tree.root, 0);        tree.root = tree.insert(tree.root, 6);        tree.root = tree.insert(tree.root, 11);        tree.root = tree.insert(tree.root, -1);        tree.root = tree.insert(tree.root, 1);        tree.root = tree.insert(tree.root, 2);         /* The constructed AVL Tree would be        9        / \        1 10        / \ \        0 5 11        / / \        -1 2 6        */        System.out.println("Preorder traversal of "+                            "constructed tree is : ");        tree.preOrder(tree.root);         tree.root = tree.deleteNode(tree.root, 10);         /* The AVL Tree after deletion of 10        1        / \        0 9        /     / \        -1 5 11        / \        2 6        */        System.out.println("");        System.out.println("Preorder traversal after "+                        "deletion of 10 :");        tree.preOrder(tree.root);    }} // This code has been contributed by Mayank Jaiswal
# Python code to delete a node in AVL tree# Generic tree node classclass TreeNode(object):    def __init__(self, val):        self.val = val        self.left = None        self.right = None        self.height = 1 # AVL tree class which supports insertion,# deletion operationsclass AVL_Tree(object):     def insert(self, root, key):                 # Step 1 - Perform normal BST        if not root:            return TreeNode(key)        elif key < root.val:            root.left = self.insert(root.left, key)        else:            root.right = self.insert(root.right, key)         # Step 2 - Update the height of the        # ancestor node        root.height = 1 + max(self.getHeight(root.left),                          self.getHeight(root.right))         # Step 3 - Get the balance factor        balance = self.getBalance(root)         # Step 4 - If the node is unbalanced,        # then try out the 4 cases        # Case 1 - Left Left        if balance > 1 and key < root.left.val:            return self.rightRotate(root)         # Case 2 - Right Right        if balance < -1 and key > root.right.val:            return self.leftRotate(root)         # Case 3 - Left Right        if balance > 1 and key > root.left.val:            root.left = self.leftRotate(root.left)            return self.rightRotate(root)         # Case 4 - Right Left        if balance < -1 and key < root.right.val:            root.right = self.rightRotate(root.right)            return self.leftRotate(root)         return root     # Recursive function to delete a node with    # given key from subtree with given root.    # It returns root of the modified subtree.    def delete(self, root, key):         # Step 1 - Perform standard BST delete        if not root:            return root         elif key < root.val:            root.left = self.delete(root.left, key)         elif key > root.val:            root.right = self.delete(root.right, key)         else:            if root.left is None:                temp = root.right                root = None                return temp             elif root.right is None:                temp = root.left                root = None                return temp             temp = self.getMinValueNode(root.right)            root.val = temp.val            root.right = self.delete(root.right,                                      temp.val)         # If the tree has only one node,        # simply return it        if root is None:            return root         # Step 2 - Update the height of the        # ancestor node        root.height = 1 + max(self.getHeight(root.left),                            self.getHeight(root.right))         # Step 3 - Get the balance factor        balance = self.getBalance(root)         # Step 4 - If the node is unbalanced,        # then try out the 4 cases        # Case 1 - Left Left        if balance > 1 and self.getBalance(root.left) >= 0:            return self.rightRotate(root)         # Case 2 - Right Right        if balance < -1 and self.getBalance(root.right) <= 0:            return self.leftRotate(root)         # Case 3 - Left Right        if balance > 1 and self.getBalance(root.left) < 0:            root.left = self.leftRotate(root.left)            return self.rightRotate(root)         # Case 4 - Right Left        if balance < -1 and self.getBalance(root.right) > 0:            root.right = self.rightRotate(root.right)            return self.leftRotate(root)         return root     def leftRotate(self, z):         y = z.right        T2 = y.left         # Perform rotation        y.left = z        z.right = T2         # Update heights        z.height = 1 + max(self.getHeight(z.left),                         self.getHeight(z.right))        y.height = 1 + max(self.getHeight(y.left),                         self.getHeight(y.right))         # Return the new root        return y     def rightRotate(self, z):         y = z.left        T3 = y.right         # Perform rotation        y.right = z        z.left = T3         # Update heights        z.height = 1 + max(self.getHeight(z.left),                          self.getHeight(z.right))        y.height = 1 + max(self.getHeight(y.left),                          self.getHeight(y.right))         # Return the new root        return y     def getHeight(self, root):        if not root:            return 0         return root.height     def getBalance(self, root):        if not root:            return 0         return self.getHeight(root.left) - self.getHeight(root.right)     def getMinValueNode(self, root):        if root is None or root.left is None:            return root         return self.getMinValueNode(root.left)     def preOrder(self, root):         if not root:            return         print("{0} ".format(root.val), end="")        self.preOrder(root.left)        self.preOrder(root.right)  myTree = AVL_Tree()root = Nonenums = [9, 5, 10, 0, 6, 11, -1, 1, 2] for num in nums:    root = myTree.insert(root, num) # Preorder Traversalprint("Preorder Traversal after insertion -")myTree.preOrder(root)print() # Deletekey = 10root = myTree.delete(root, key) # Preorder Traversalprint("Preorder Traversal after deletion -")myTree.preOrder(root)print() # This code is contributed by Ajitesh Pathak
// C# program for deletion in AVL Treeusing System; public class Node{    public int key, height;    public Node left, right;     public Node(int d)    {        key = d;        height = 1;    }} public class AVLTree{    Node root;     // A utility function to get height of the tree    int height(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;    }     // A utility function to right    // rotate subtree rooted with y    // See the diagram given above.    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.    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    int getBalance(Node N)    {        if (N == null)            return 0;        return height(N.left) - height(N.right);    }     Node insert(Node node, int key)    {        /* 1. Perform the normal BST rotation */        if (node == null)            return (new Node(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        Wunbalanced */        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. */    Node minValueNode(Node node)    {        Node current = node;         /* loop down to find the leftmost leaf */        while (current.left != null)        current = current.left;         return current;    }     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 = null;                if (temp == root.left)                    temp = root.right;                else                    temp = root.left;                 // 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                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(Node node)    {        if (node != null)        {            Console.Write(node.key + " ");            preOrder(node.left);            preOrder(node.right);        }    }     // Driver code    public static void Main()    {        AVLTree tree = new AVLTree();         /* Constructing tree given in the above figure */        tree.root = tree.insert(tree.root, 9);        tree.root = tree.insert(tree.root, 5);        tree.root = tree.insert(tree.root, 10);        tree.root = tree.insert(tree.root, 0);        tree.root = tree.insert(tree.root, 6);        tree.root = tree.insert(tree.root, 11);        tree.root = tree.insert(tree.root, -1);        tree.root = tree.insert(tree.root, 1);        tree.root = tree.insert(tree.root, 2);         /* The constructed AVL Tree would be        9        / \        1 10        / \ \        0 5 11        / / \        -1 2 6        */        Console.WriteLine("Preorder traversal of "+                            "constructed tree is : ");        tree.preOrder(tree.root);         tree.root = tree.deleteNode(tree.root, 10);         /* The AVL Tree after deletion of 10        1        / \        0 9        /     / \        -1 5 11        / \        2 6        */        Console.WriteLine("");        Console.WriteLine("Preorder traversal after "+                        "deletion of 10 :");        tree.preOrder(tree.root);    }} /* This code contributed by PrinciRaj1992 */
<script>         // JavaScript program for deletion in AVL Tree         class Node    {        constructor(d) {           this.left = null;           this.right = null;           this.key = d;           this.height = 1;        }    }         let root;       // A utility function to get height of the tree    function height(N)    {        if (N == null)            return 0;        return N.height;    }       // A utility function to get maximum of two integers    function max(a, b)    {        return (a > b) ? a : b;    }       // A utility function to right rotate subtree rooted with y    // See the diagram given above.    function rightRotate(y)    {        let x = y.left;        let 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.    function leftRotate(x)    {        let y = x.right;        let 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    function getBalance(N)    {        if (N == null)            return 0;        return height(N.left) - height(N.right);    }       function insert(node, key)    {        /* 1. Perform the normal BST rotation */        if (node == null)            return (new Node(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        Wunbalanced */        let 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. */    function minValueNode(node)    {        let current = node;           /* loop down to find the leftmost leaf */        while (current.left != null)            current = current.left;           return current;    }       function deleteNode(root, 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))            {                let temp = null;                if (temp == root.left)                    temp = root.right;                else                    temp = root.left;                   // 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)                let 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)        let 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    function preOrder(node)    {        if (node != null)        {            document.write(node.key + " ");            preOrder(node.left);            preOrder(node.right);        }    }         /* 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          */    document.write(    "Preorder traversal of the constructed AVL tree is : " +    "</br>");    preOrder(root);     root = deleteNode(root, 10);     /* The AVL Tree after deletion of 10          1          / \          0 9          /     / \          -1 5 11          / \          2 6          */    document.write("</br>");    document.write("Preorder traversal after "+                       "deletion of 10 :" + "</br>");    preOrder(root);   </script>
Output:  
Preorder traversal of the constructed AVL tree is 
9 1 0 -1 5 2 6 10 11 
Preorder traversal after deletion of 10 
1 0 -1 9 5 2 6 11 
Time Complexity: The rotation operations (left and right rotate) take constant time as only few pointers are being changed there. Updating the height and getting the balance factor also take constant time. So the time complexity of AVL delete remains same as BST delete which is O(h) where h is height of the tree. Since AVL tree is balanced, the height is O(Logn). So time complexity of AVL delete is O(Log n). 
Advantages Of AVL Trees
It is always height balanced
Height Never Goes Beyond LogN, where N is the number of nodes
It give better search than compared to binary search tree
It has self balancing capabilities
Summary of AVL Trees
These are self-balancing binary search trees.
Balancing Factor ranges -1, 0, and +1.
When balancing factor goes beyond the range require rotations to be performed
Insert, delete, and search time is O(log N).
AVL tree are mostly used where search is more frequent compared to insert and delete operations.
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References: https://www.cs.purdue.edu/homes/ayg/CS251/slides/chap7b.pdf IITD Video Lecture on AVL Tree Insertion and DeletionPlease write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
 
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