Count greater nodes in AVL tree

• Difficulty Level : Hard
• Last Updated : 24 Oct, 2017

In this article we will see that how to calculate number of elements which are greater than given value in AVL tree.
Examples:

Input : x = 5
Root of below AVL tree
9
/  \
1    10
/  \     \
0    5     11
/    /  \
-1   2    6
Output : 4

Explanation: there are 4 values which are
greater than 5 in AVL tree which are 6, 9,
10 and 11.

Prerequisites :

1. We maintain an extra field ‘desc‘ for storing the number of descendant nodes for every node. Like for above example node having value 5 has a desc field value equal to 2.

2. for calculating the number of nodes which are greater than given value we simply traverse the tree. While traversing three cases can occur-

I Case- x(given value) is greater than the value of current node. So, we go to the right child of the current node.
II Case- x is lesser than the value of current node. we increase the current count by number of successors of the right child of the current node and then again add two to the current count(one for the current node and one for the right child.). In this step first, we make sure that right child exists or not. Then we move to left child of current node.
III Case-x is equal to the value of current node. In this case we add the value of desc field of right child of current node to current count and then add one to it (for counting right child). Also in this case we see that right child exists or not.

Calculating values of desc field

1. Insertion – When we insert a node we increment one to child field of every predeccesor of the new node. In the leftRotate and rightRotate functions we make appropriate changes in the value of child fields of nodes.
2. Deletion – When we delete a node then we decrement one from every prdeccessor node of deleted node. Again, In the leftRotate and rightRotate functions we make appropriate changes in the value of child fields of nodes.
 // C program to find number of elements// greater than a given value in AVL#include #include struct Node {    int key;    struct Node* left, *right;    int height;    int desc;};  int 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;}  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; // initially added at leaf    node->desc = 0;    return (node);}  // A utility function to right rotate subtree// rooted with ystruct 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;      // calculate the number of children of x and y    // which are changed due to rotation.    int val = (T2 != NULL) ? T2->desc : -1;    y->desc = y->desc - (x->desc + 1) + (val + 1);    x->desc = x->desc - (val + 1) + (y->desc + 1);      return x;}  // A utility function to left rotate subtree rooted// with xstruct 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;      // calculate the number of children of x and y    // which are changed due to rotation.    int val = (T2 != NULL) ? T2->desc : -1;    x->desc = x->desc - (y->desc + 1) + (val + 1);    y->desc = y->desc - (val + 1) + (x->desc + 1);      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);        node->desc++;    }      else if (key > node->key) {        node->right = insert(node->right, key);        node->desc++;    }      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 node becomes unbalanced, 4 cases arise      // 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);        root->desc = root->desc - 1;    }      // 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);        root->desc = root->desc - 1;    }      // 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;                free(temp);              }             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);            root->desc = root->desc - 1;        }    }      // 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, 4 cases arise      // 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.void preOrder(struct Node* root){    if (root != NULL) {        printf("%d ", root->key);        preOrder(root->left);        preOrder(root->right);    }}  // Returns count ofint CountGreater(struct Node* root, int x){    int res = 0;      // Search for x. While searching, keep    // updating res if x is greater than    // current node.    while (root != NULL) {          int desc = (root->right != NULL) ?                    root->right->desc : -1;          if (root->key > x) {            res = res + desc + 1 + 1;            root = root->left;        } else if (root->key < x)            root = root->right;        else {            res = res + desc + 1;            break;        }    }    return res;}  /* Driver program to test above function*/int main(){    struct Node* root = NULL;    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);    printf("\nNumber of elements greater than 9 are %d",           CountGreater(root, 9));      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);    printf("\nNumber of elements greater than 9 are %d",           CountGreater(root, 9));    return 0;}

Output:

Preorder traversal of the constructed AVL tree is
9 1 0 -1 5 2 6 10 11
Number of elements greater than 9 are 2
Preorder traversal after deletion of 10
1 0 -1 9 5 2 6 11
Number of elements greater than 9 are 1

Time Complexity: Time complexity of CountGreater function is O(log(n)) where n is number of nodes in avl tree, as we are basically searching for the given number in avl which takes O(log(n)) time.

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