Consulte la publicación a continuación antes de leer sobre el manejo de árboles AVL de duplicados.
¿Cómo manejar los duplicados en el árbol de búsqueda binaria?
Esto es para aumentar el Node del árbol AVL para almacenar el recuento junto con campos regulares como punteros clave, izquierdo y derecho.
La inserción de las claves 12, 10, 20, 9, 11, 10, 12, 12 en un árbol de búsqueda binario vacío crearía lo siguiente.
12(3)
/ \
10(2) 20(1)
/ \
9(1) 11(1)
El recuento de una clave se muestra entre paréntesis
A continuación se muestra la implementación del árbol AVL normal con el recuento de cada clave. Este código básicamente se toma del código para insertar y eliminar en el árbol AVL . Los cambios realizados para el manejo de duplicados están resaltados, el resto del código es el mismo.
Lo importante a tener en cuenta es que los cambios son muy similares a los cambios simples del árbol de búsqueda binaria.
C++
// C++ program of AVL tree that
// handles duplicates
#include <bits/stdc++.h>
using namespace std;
// 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 NULL;
}
// 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) {
cout << 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);
cout <<"Pre order traversal of the constructed AVL tree is \n";
preOrder(root);
cout <<"\nPre order traversal after deletion of 9 \n";
preOrder(root);
return 0;
}
// this code is contributed by shivanisinghss2110
C
// C++ program of AVL tree that
// handles duplicates
#include <stdio.h>
#include <stdlib.h>
// 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 NULL;
}
// 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;
/* 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);
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,
// 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) {
Console.Write(root.key + "(" + root.count + ") ");
preOrder(root.left);
preOrder(root.right);
}
}
// Driver Code
static public void Main(String[] args)
{
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);
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
Python3
# Python code to delete a node in AVL tree
# Generic tree node class
class TreeNode():
def __init__(self, val):
self.count = 1 # assigning count variable so that during insertion in will be incremented for duplicate values
# and during deletion, it will be decremented if has multiple copies.
self.height = 1
self.val = val
self.left = None
self.right = None
# only insertion and deletion will be affected. if multiple copies are there, entry(count) will be printed during traversal.
# AVL tree class which supports insertion,
# deletion operations
class AVL_Tree(object):
def insert(self, root, key):
# Step 1 - Perform normal BST
if not root:
return TreeNode(key)
else if key < root.val:
root.left = self.insert(root.left, key)
else if key > root.val:
root.right = self.insert(root.right, key)
else:
root.count += 1 # incrementing count if same entry is inserted.
# 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
else if key < root.val:
root.left = self.delete(root.left, key)
else if key > root.val:
root.right = self.delete(root.right, key)
else:
if root.count > 1: # if count is more than one i.e multiple copies are there
root.count -= 1 # just decrement count
return root # so that one copy will be deleted and return
if root.left is None:
temp = root.right
root = None
return temp
else if 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("{}({}) ".format(root.val, root.count), end="")
self.preOrder(root.left)
self.preOrder(root.right)
myTree = AVL_Tree()
root = None
nums = [9, 5, 10, 5, 9, 7, 17]
for num in nums:
root = myTree.insert(root, num)
# Preorder Traversal
print("Preorder Traversal after insertion -")
myTree.preOrder(root)
print()
# Delete
key = 10
root = myTree.delete(root, key)
key = 10
root = myTree.delete(root, key)
key = -1
root = myTree.delete(root, key)
key = 0
root = myTree.delete(root, key)
# Preorder Traversal
print("Preorder Traversal after deletion -")
myTree.preOrder(root)
print()
# This code is contributed by Ajitesh Pathak
Javascript
<script>
// JavaScript program of AVL tree that handles duplicates
class Node
{
constructor() {
this.left;
this.right;
this.key;
this.height;
this.count;
}
}
// 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;
}
/* Helper function that allocates a new
node with the given key and
null left and right pointers. */
function newNode(key)
{
let 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.
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 (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 */
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 {
// 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)) {
let temp = root.left != null ? root.left : root.right;
// No child case
if (temp == null) {
temp = root;
root = null;
}
else // One child case
// Copy the contents of the non-empty child
root = temp;
}
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 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)
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(root)
{
if (root != null) {
document.write(root.key + "(" + root.count + ") ");
preOrder(root.left);
preOrder(root.right);
}
}
let 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);
document.write(
"Pre order traversal of the constructed AVL tree is " +
"</br>");
preOrder(root);
deleteNode(root, 9);
document.write(
"</br>" + "Pre order traversal after deletion of 9 " +
"</br>");
preOrder(root);
</script>
Producción
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(2) 5(2) 7(1) 10(1) 17(1)
Gracias a Rounaq Jhunjhunu Wala por compartir el código inicial. Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA