计算 AVL 树中更大的节点
原文:https://www . geesforgeks . org/count-greater-nodes-in-AVL-tree/
在本文中我们将看到如何计算 AVL 树中大于给定值的元素个数。 例子:
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.
先决条件:
1.我们维护一个额外的字段“ desc ”，用于存储每个节点的后代节点数。像上面的例子一样，值为 5 的节点具有等于 2 的 desc 场值。
2.为了计算大于给定值的节点数，我们简单地遍历树。当遍历三个案例时-
I Case- x(给定值)大于当前节点的值。因此，我们转到当前节点的右子节点。 案例二- x 小于当前节点的值。我们通过当前节点的右子节点的后继数增加当前计数，然后再次将两个加到当前计数中(一个用于当前节点，一个用于右子节点。).在这一步中，首先，我们要确定正确的孩子是否存在。然后我们移动到当前节点的左子节点。 III 情况- x 等于当前节点的值。在这种情况下，我们将当前节点的右子节点的 desc 字段的值添加到当前计数中，然后再添加一个(用于计数右子节点)。同样在这种情况下，我们看到正确的孩子存在与否。
计算 desc 场的数值
- 插入–当我们插入一个节点时，我们将新节点的每个预解码器的子字段递增 1。在左旋转和右旋转函数中，我们对节点的子字段值进行了适当的更改。
- 删除–当我们删除一个节点时，我们从被删除节点的每个处理器节点中减少一个。同样，在左旋转和右旋转函数中，我们对节点的子字段的值进行了适当的更改。
// C program to find number of elements
// greater than a given value in AVL
#include <stdio.h>
#include <stdlib.h>
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 integers
int 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 y
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;
// 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 x
struct Node* leftRotate(struct Node* x)
{
struct Node* y = x->right;
struct Node* T2 = y->left;
// Perform rotation
y->left = x;
x->right = T2;
// Update heights
x->height = max(height(x->left), height(x->right)) + 1;
y->height = max(height(y->left), height(y->right)) + 1;
// 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 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);
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 of
int 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;
}
输出:
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
时间复杂度:count 的时间复杂度更大的函数是 O(log(n))，其中 n 是 avl 树中的节点数，因为我们基本上是在 avl 中搜索给定的需要 O(log(n))时间的数。
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