En un árbol de búsqueda binaria (BST), todas las claves del subárbol izquierdo de una clave deben ser más pequeñas y todas las claves del subárbol derecho deben ser mayores. Entonces, un árbol de búsqueda binario , por definición, tiene claves distintas.
¿Cómo permitir duplicados donde cada inserción inserta una clave más con un valor y cada eliminación elimina una ocurrencia?
Una solución simple es permitir las mismas teclas en el lado derecho (también podemos elegir el lado izquierdo). Por ejemplo, considere la inserción de las claves 12, 10, 20, 9, 11, 10, 12, 12 en un árbol de búsqueda binario vacío
12
/ \
10 20
/ \ /
9 11 12
/ \
10 12
Una mejor solución es aumentar cada Node del árbol 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)
Count of a key is shown in bracket
Este enfoque tiene las siguientes ventajas sobre el enfoque simple anterior.
- La altura del árbol es pequeña independientemente del número de duplicados. Tenga en cuenta que la mayoría de las operaciones de BST (búsqueda, inserción y eliminación) tienen una complejidad de tiempo como O(h), donde h es la altura de BST. Entonces, si podemos mantener la altura pequeña, obtenemos la ventaja de una menor cantidad de comparaciones clave.
- Buscar, Insertar y Eliminar se vuelven más fáciles de hacer. Podemos usar los mismos algoritmos de inserción, búsqueda y eliminación con pequeñas modificaciones (ver el código a continuación).
- Este enfoque también es adecuado para BST autoequilibrados ( AVL Tree , Red-Black Tree , etc.). Estos árboles involucran rotaciones, y una rotación puede violar la propiedad BST de solución simple ya que una misma tecla puede estar en el lado izquierdo o derecho después de la rotación.
A continuación se muestra la implementación del árbol de búsqueda binaria normal con recuento de cada clave. Este código básicamente se toma del código para insertar y eliminar en BST . Los cambios realizados para el manejo de duplicados están resaltados, el resto del código es el mismo.
C++
// C++ program to implement basic operations
// (search, insert and delete) on a BST that
// handles duplicates by storing count with
// every node
#include<bits/stdc++.h>
using namespace std;
struct node
{
int key;
int count;
struct node *left, *right;
};
// A utility function to create a new BST node
struct node *newNode(int item)
{
struct node *temp = (struct node *)malloc(sizeof(struct node));
temp->key = item;
temp->left = temp->right = NULL;
temp->count = 1;
return temp;
}
// A utility function to do inorder traversal of BST
void inorder(struct node *root)
{
if (root != NULL)
{
inorder(root->left);
cout << root->key << "("
<< root->count << ") ";
inorder(root->right);
}
}
/* A utility function to insert a new
node with given key in BST */
struct node* insert(struct node* node, int key)
{
/* If the tree is empty, return a new node */
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);
/* 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;
}
/* Given a binary search tree and a key,
this function deletes a given key and
returns root of modified tree */
struct node* deleteNode(struct node* root,
int key)
{
// base case
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
else
{
// If key is present more than once,
// simply decrement count and return
if (root->count > 1)
{
(root->count)--;
return root;
}
// ElSE, delete the node
// node with only one child or no child
if (root->left == NULL)
{
struct node *temp = root->right;
free(root);
return temp;
}
else if (root->right == NULL)
{
struct node *temp = root->left;
free(root);
return temp;
}
// node with two children: Get the inorder
// successor (smallest in the right subtree)
struct node* temp = minValueNode(root->right);
// Copy the inorder successor's
// content to this node
root->key = temp->key;
root->count = temp->count;
// To ensure successor gets deleted by
// deleteNode call, set count to 0.
temp->count = 0;
// Delete the inorder successor
root->right = deleteNode(root->right,
temp->key);
}
return root;
}
// Driver Code
int main()
{
/* Let us create following BST
12(3)
/ \
10(2) 20(1)
/ \
9(1) 11(1) */
struct node *root = NULL;
root = insert(root, 12);
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 9);
root = insert(root, 11);
root = insert(root, 10);
root = insert(root, 12);
root = insert(root, 12);
cout << "Inorder traversal of the given tree "
<< endl;
inorder(root);
cout << "\nDelete 20\n";
root = deleteNode(root, 20);
cout << "Inorder traversal of the modified tree \n";
inorder(root);
cout << "\nDelete 12\n" ;
root = deleteNode(root, 12);
cout << "Inorder traversal of the modified tree \n";
inorder(root);
cout << "\nDelete 9\n";
root = deleteNode(root, 9);
cout << "Inorder traversal of the modified tree \n";
inorder(root);
return 0;
}
// This code is contributed by Akanksha Rai
C
// C program to implement basic operations (search, insert and delete)
// on a BST that handles duplicates by storing count with every node
#include<stdio.h>
#include<stdlib.h>
struct node
{
int key;
int count;
struct node *left, *right;
};
// A utility function to create a new BST node
struct node *newNode(int item)
{
struct node *temp = (struct node *)malloc(sizeof(struct node));
temp->key = item;
temp->left = temp->right = NULL;
temp->count = 1;
return temp;
}
// A utility function to do inorder traversal of BST
void inorder(struct node *root)
{
if (root != NULL)
{
inorder(root->left);
printf("%d(%d) ", root->key, root->count);
inorder(root->right);
}
}
/* A utility function to insert a new node with given key in BST */
struct node* insert(struct node* node, int key)
{
/* If the tree is empty, return a new node */
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);
/* 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;
}
/* Given a binary search tree and a key, this function
deletes a given key and returns root of modified tree */
struct node* deleteNode(struct node* root, int key)
{
// base case
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
else
{
// If key is present more than once, simply decrement
// count and return
if (root->count > 1)
{
(root->count)--;
return root;
}
// ElSE, delete the node
// node with only one child or no child
if (root->left == NULL)
{
struct node *temp = root->right;
free(root);
return temp;
}
else if (root->right == NULL)
{
struct node *temp = root->left;
free(root);
return temp;
}
// node with two children: Get the inorder successor (smallest
// in the right subtree)
struct node* temp = minValueNode(root->right);
// Copy the inorder successor's content to this node
root->key = temp->key;
root->count = temp->count;
// Delete the inorder successor
root->right = deleteNode(root->right, temp->key);
}
return root;
}
// Driver Program to test above functions
int main()
{
/* Let us create following BST
12(3)
/ \
10(2) 20(1)
/ \
9(1) 11(1) */
struct node *root = NULL;
root = insert(root, 12);
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 9);
root = insert(root, 11);
root = insert(root, 10);
root = insert(root, 12);
root = insert(root, 12);
printf("Inorder traversal of the given tree \n");
inorder(root);
printf("\nDelete 20\n");
root = deleteNode(root, 20);
printf("Inorder traversal of the modified tree \n");
inorder(root);
printf("\nDelete 12\n");
root = deleteNode(root, 12);
printf("Inorder traversal of the modified tree \n");
inorder(root);
printf("\nDelete 9\n");
root = deleteNode(root, 9);
printf("Inorder traversal of the modified tree \n");
inorder(root);
return 0;
}
Java
// Java program to implement basic operations
// (search, insert and delete) on a BST that
// handles duplicates by storing count with
// every node
class GFG
{
static class node
{
int key;
int count;
node left, right;
};
// A utility function to create a new BST node
static node newNode(int item)
{
node temp = new node();
temp.key = item;
temp.left = temp.right = null;
temp.count = 1;
return temp;
}
// A utility function to do inorder traversal of BST
static void inorder(node root)
{
if (root != null)
{
inorder(root.left);
System.out.print(root.key + "(" +
root.count + ") ");
inorder(root.right);
}
}
/* A utility function to insert a new
node with given key in BST */
static node insert(node, int key)
{
/* If the tree is empty, return a new node */
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);
/* 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;
}
/* Given a binary search tree and a key,
this function deletes a given key and
returns root of modified tree */
static node deleteNode(node root, int key)
{
// base case
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
else
{
// If key is present more than once,
// simply decrement count and return
if (root.count > 1)
{
(root.count)--;
return root;
}
// ElSE, delete the node
// node with only one child or no child
if (root.left == null)
{
node temp = root.right;
root=null;
return temp;
}
else if (root.right == null)
{
node temp = root.left;
root = null;
return temp;
}
// node with two children: Get the inorder
// successor (smallest in the right subtree)
node temp = minValueNode(root.right);
// Copy the inorder successor's
// content to this node
root.key = temp.key;
root.count = temp.count;
// Delete the inorder successor
root.right = deleteNode(root.right,
temp.key);
}
return root;
}
// Driver Code
public static void main(String[] args)
{
/* Let us create following BST
12(3)
/ \
10(2) 20(1)
/ \
9(1) 11(1) */
node root = null;
root = insert(root, 12);
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 9);
root = insert(root, 11);
root = insert(root, 10);
root = insert(root, 12);
root = insert(root, 12);
System.out.print("Inorder traversal of " +
"the given tree " + "\n");
inorder(root);
System.out.print("\nDelete 20\n");
root = deleteNode(root, 20);
System.out.print("Inorder traversal of " +
"the modified tree \n");
inorder(root);
System.out.print("\nDelete 12\n");
root = deleteNode(root, 12);
System.out.print("Inorder traversal of " +
"the modified tree \n");
inorder(root);
System.out.print("\nDelete 9\n");
root = deleteNode(root, 9);
System.out.print("Inorder traversal of " +
"the modified tree \n");
inorder(root);
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 program to implement basic operations
# (search, insert and delete) on a BST that handles
# duplicates by storing count with every node
# A utility function to create a new BST node
class newNode:
# Constructor to create a new node
def __init__(self, data):
self.key = data
self.count = 1
self.left = None
self.right = None
# A utility function to do inorder
# traversal of BST
def inorder(root):
if root != None:
inorder(root.left)
print(root.key,"(", root.count,")",
end = " ")
inorder(root.right)
# A utility function to insert a new node
# with given key in BST
def insert(node, key):
# If the tree is empty, return a new node
if node == None:
k = newNode(key)
return k
# If key already exists in BST, increment
# count and return
if key == node.key:
(node.count) += 1
return node
# Otherwise, recur down the tree
if key < node.key:
node.left = insert(node.left, key)
else:
node.right = insert(node.right, key)
# 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.
def minValueNode(node):
current = node
# loop down to find the leftmost leaf
while current.left != None:
current = current.left
return current
# Given a binary search tree and a key,
# this function deletes a given key and
# returns root of modified tree
def deleteNode(root, key):
# base case
if root == None:
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
else:
# If key is present more than once,
# simply decrement count and return
if root.count > 1:
root.count -= 1
return root
# ElSE, delete the node with
# only one child or no child
if root.left == None:
temp = root.right
return temp
else if root.right == None:
temp = root.left
return temp
# node with two children: Get the inorder
# successor (smallest in the right subtree)
temp = minValueNode(root.right)
# Copy the inorder successor's content
# to this node
root.key = temp.key
root.count = temp.count
# Delete the inorder successor
root.right = deleteNode(root.right, temp.key)
return root
# Driver Code
if __name__ == '__main__':
# Let us create following BST
# 12(3)
# / \
# 10(2) 20(1)
# / \
# 9(1) 11(1)
root = None
root = insert(root, 12)
root = insert(root, 10)
root = insert(root, 20)
root = insert(root, 9)
root = insert(root, 11)
root = insert(root, 10)
root = insert(root, 12)
root = insert(root, 12)
print("Inorder traversal of the given tree")
inorder(root)
print()
print("Delete 20")
root = deleteNode(root, 20)
print("Inorder traversal of the modified tree")
inorder(root)
print()
print("Delete 12")
root = deleteNode(root, 12)
print("Inorder traversal of the modified tree")
inorder(root)
print()
print("Delete 9")
root = deleteNode(root, 9)
print("Inorder traversal of the modified tree")
inorder(root)
# This code is contributed by PranchalK
C#
// C# program to implement basic operations
// (search, insert and delete) on a BST that
// handles duplicates by storing count with
// every node
using System;
class GFG
{
public class node
{
public int key;
public int count;
public node left, right;
};
// A utility function to create
// a new BST node
static node newNode(int item)
{
node temp = new node();
temp.key = item;
temp.left = temp.right = null;
temp.count = 1;
return temp;
}
// A utility function to do inorder
// traversal of BST
static void inorder(node root)
{
if (root != null)
{
inorder(root.left);
Console.Write(root.key + "(" +
root.count + ") ");
inorder(root.right);
}
}
/* A utility function to insert a new
node with given key in BST */
static node insert(node, int key)
{
/* If the tree is empty,
return a new node */
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);
/* 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;
}
/* Given a binary search tree and a key,
this function deletes a given key and
returns root of modified tree */
static node deleteNode(node root, int key)
{
// base case
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
else
{
// If key is present more than once,
// simply decrement count and return
if (root.count > 1)
{
(root.count)--;
return root;
}
// ElSE, delete the node
node temp = null;
// node with only one child or no child
if (root.left == null)
{
temp = root.right;
root = null;
return temp;
}
else if (root.right == null)
{
temp = root.left;
root = null;
return temp;
}
// node with two children: Get the inorder
// successor (smallest in the right subtree)
temp = minValueNode(root.right);
// Copy the inorder successor's
// content to this node
root.key = temp.key;
root.count = temp.count;
// Delete the inorder successor
root.right = deleteNode(root.right,
temp.key);
}
return root;
}
// Driver Code
public static void Main(String[] args)
{
/* Let us create following BST
12(3)
/ \
10(2) 20(1)
/ \
9(1) 11(1) */
node root = null;
root = insert(root, 12);
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 9);
root = insert(root, 11);
root = insert(root, 10);
root = insert(root, 12);
root = insert(root, 12);
Console.Write("Inorder traversal of " +
"the given tree " + "\n");
inorder(root);
Console.Write("\nDelete 20\n");
root = deleteNode(root, 20);
Console.Write("Inorder traversal of " +
"the modified tree \n");
inorder(root);
Console.Write("\nDelete 12\n");
root = deleteNode(root, 12);
Console.Write("Inorder traversal of " +
"the modified tree \n");
inorder(root);
Console.Write("\nDelete 9\n");
root = deleteNode(root, 9);
Console.Write("Inorder traversal of " +
"the modified tree \n");
inorder(root);
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// JavaScript program to implement basic operations
// (search, insert and delete) on a BST that
// handles duplicates by storing count with
// every node
class node {
constructor() {
this.key = 0;
this.count = 0;
this.left = null;
this.right = null;
}
}
// A utility function to create a new BST node
function newNode(item) {
var temp = new node();
temp.key = item;
temp.left = temp.right = null;
temp.count = 1;
return temp;
}
// A utility function to do inorder
// traversal of BST
function inorder( root) {
if (root != null) {
inorder(root.left);
document.write(root.key +
"(" + root.count + ") ");
inorder(root.right);
}
}
/*
* A utility function to insert a new
node with given key in BST
*/
function insert( node , key) {
/* If the tree is empty, return a new node */
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);
/* 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) {
var current = node;
/* loop down to find the leftmost leaf */
while (current.left != null)
current = current.left;
return current;
}
/*
* Given a binary search tree and a key,
this function deletes a given key and
* returns root of modified tree
*/
function deleteNode( root , key) {
// base case
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
else {
// If key is present more than once,
// simply decrement count and return
if (root.count > 1) {
(root.count)--;
return root;
}
// ElSE, delete the node
// node with only one child or no child
if (root.left == null) {
var temp = root.right;
root = null;
return temp;
} else if (root.right == null) {
var temp = root.left;
root = null;
return temp;
}
// node with two children: Get the inorder
// successor (smallest in the right subtree)
var temp = minValueNode(root.right);
// Copy the inorder successor's
// content to this node
root.key = temp.key;
root.count = temp.count;
// Delete the inorder successor
root.right = deleteNode(root.right, temp.key);
}
return root;
}
// Driver Code
/*
* Let us create following BST
12(3) / \ 10(2) 20(1) / \ 9(1) 11(1)
*/
var root = null;
root = insert(root, 12);
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 9);
root = insert(root, 11);
root = insert(root, 10);
root = insert(root, 12);
root = insert(root, 12);
document.write("Inorder traversal of "
+ "the given tree " + "<br/>");
inorder(root);
document.write("<br/>Delete 20<br/>");
root = deleteNode(root, 20);
document.write("Inorder traversal of "
+ "the modified tree <br/>");
inorder(root);
document.write("<br/>Delete 12<br/>");
root = deleteNode(root, 12);
document.write("Inorder traversal of "
+ "the modified tree <br/>");
inorder(root);
document.write("<br/>Delete 9<br/>");
root = deleteNode(root, 9);
document.write("Inorder traversal of "
+ "the modified tree <br/>");
inorder(root);
// This code contributed by aashish1995
</script>
Producción
Inorder traversal of the given tree 9(1) 10(2) 11(1) 12(3) 20(1) Delete 20 Inorder traversal of the modified tree 9(1) 10(2) 11(1) 12(3) Delete 12 Inorder traversal of the modified tree 9(1) 10(2) 11(1) 12(2) Delete 9 Inorder traversal of the modified tree 10(2) 11(1) 12(2)
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA