Dada una array arr[] de enteros, la tarea es eliminar los duplicados de la array dada.
Ejemplos :
Input: arr[] = {1, 2, 3, 2, 5, 4, 4} Output: arr[] = {1, 2, 3, 4, 5} Input: arr[] = {127, 234, 127, 654, 355, 789, 355, 355, 999, 654} Output: arr[] = {127, 234, 355, 654, 789, 999}
Los duplicados en la array se pueden eliminar mediante el árbol de búsqueda binaria . La idea es crear un árbol de búsqueda binario utilizando los elementos de la array con la condición de que el primer elemento se tome como el elemento raíz (padre) y cuando aparezca el elemento «menor» que la raíz, se convierta en el elemento secundario izquierdo y el elemento » mayor” que root se convierte en el hijo derecho de la raíz. Dado que no existe ninguna condición para «igual», los duplicados se eliminan automáticamente cuando formamos un árbol de búsqueda binaria a partir de los elementos de la array.
Para la array, arr[] = {1, 2, 3, 2, 5, 4, 4}
BST será:
Acercarse:
- Forme BST usando los elementos de la array
- Muestre los elementos usando cualquier método Tree Traversal .
A continuación se muestra la implementación del enfoque anterior.
C++
// C++ Program of above implementation #include <iostream> using namespace std; // Struct declaration struct Node { int data; struct Node* left; struct Node* right; }; // Node creation struct Node* newNode(int data) { struct Node* nn = new Node; nn->data = data; nn->left = NULL; nn->right = NULL; return nn; } // Function to insert data in BST struct Node* insert(struct Node* root, int data) { if (root == NULL) return newNode(data); else { if (data < root->data) root->left = insert(root->left, data); if (data > root->data) root->right = insert(root->right, data); return root; } } // InOrder function to display value of array // in sorted order void inOrder(struct Node* root) { if (root == NULL) return; else { inOrder(root->left); cout << root->data << " "; inOrder(root->right); } } // Driver code int main() { int arr[] = { 1, 2, 3, 2, 5, 4, 4 }; // Finding size of array arr[] int n = sizeof(arr) / sizeof(arr[0]); struct Node* root = NULL; for (int i = 0; i < n; i++) { // Insert element of arr[] in BST root = insert(root, arr[i]); } // Inorder Traversal to print nodes of Tree inOrder(root); return 0; } // This code is contributed by shivanisingh
C
// C Program of above implementation #include <stdio.h> #include <stdlib.h> // Struct declaration struct Node { int data; struct Node* left; struct Node* right; }; // Node creation struct Node* newNode(int data) { struct Node* nn = (struct Node*)(malloc(sizeof(struct Node))); nn->data = data; nn->left = NULL; nn->right = NULL; return nn; } // Function to insert data in BST struct Node* insert(struct Node* root, int data) { if (root == NULL) return newNode(data); else { if (data < root->data) root->left = insert(root->left, data); if (data > root->data) root->right = insert(root->right, data); return root; } } // InOrder function to display value of array // in sorted order void inOrder(struct Node* root) { if (root == NULL) return; else { inOrder(root->left); printf("%d ", root->data); inOrder(root->right); } } // Driver code int main() { int arr[] = { 1, 2, 3, 2, 5, 4, 4 }; // Finding size of array arr[] int n = sizeof(arr) / sizeof(arr[0]); struct Node* root = NULL; for (int i = 0; i < n; i++) { // Insert element of arr[] in BST root = insert(root, arr[i]); } // Inorder Traversal to print nodes of Tree inOrder(root); return 0; }
Java
// Java implementation of the approach import java.util.Scanner; // Node declaration class Node { int data; public Node left; public Node right; Node(int data) { this.data = data; left = right = null; } } class GFG { // Function to insert data in BST public static Node insert(Node root, int data) { if (root == null) return new Node(data); if (data < root.data) root.left = insert(root.left, data); if (data > root.data) root.right = insert(root.right, data); return root; } // InOrder function to display value of array // in sorted order public static void inOrder(Node root) { if (root == null) return; inOrder(root.left); System.out.print(root.data+" "); inOrder(root.right); } // Driver Code public static void main(String []args){ int arr[] = { 1, 2, 3, 2, 5, 4, 4 }; // Finding size of array arr[] int n = arr.length; Node root = null; for (int i = 0; i < n; i++) { // Insert element of arr[] in BST root = insert(root,arr[i]); } // Inorder Traversal to print nodes of Tree inOrder(root); } } // This code is contributed by anishma
Python3
# Python3 implementation of the approach # Binary tree node consists of data, a # pointer to the left child and a # pointer to the right child class newNode : def __init__(self,data) : self.data = data; self.left = None; self.right = None; # Function to insert data in BST def insert(root, data) : if (root == None) : return newNode(data); else : if (data < root.data) : root.left = insert(root.left, data); if (data > root.data) : root.right = insert(root.right, data); return root; # InOrder function to display value of array # in sorted order def inOrder(root) : if (root == None) : return; else : inOrder(root.left); print(root.data, end = " "); inOrder(root.right); # Driver code if __name__ == "__main__" : arr = [ 1, 2, 3, 2, 5, 4, 4 ]; # Finding size of array arr[] n = len(arr); root = None; for i in range(n) : # Insert element of arr[] in BST root = insert(root, arr[i]); # Inorder Traversal to print nodes of Tree inOrder(root); # This code is contributed by AnkitRai01
C#
// C# program of above implementation using System; // Node declaration public class Node { public int data; public Node left; public Node right; public Node(int data) { this.data = data; left = right = null; } } class GFG{ // Function to insert data in BST public static Node insert(Node root, int data) { if (root == null) return new Node(data); if (data < root.data) root.left = insert(root.left, data); if (data > root.data) root.right = insert(root.right, data); return root; } // InOrder function to display value of array // in sorted order public static void inOrder(Node root) { if (root == null) return; inOrder(root.left); Console.Write(root.data + " "); inOrder(root.right); } // Driver Code static void Main() { int[] arr = { 1, 2, 3, 2, 5, 4, 4 }; // Finding size of array arr[] int n = arr.Length; Node root = null; for(int i = 0; i < n; i++) { // Insert element of arr[] in BST root = insert(root, arr[i]); } // Inorder Traversal to print nodes of Tree inOrder(root); } } // This code is contributed by divyeshrabadiya07
Javascript
<script> // JavaScript program of above implementation // Node declaration class Node { constructor(data) { this.data = data; this.left = null; this.right = null; } } // Function to insert data in BST function insert(root, data) { if (root == null) return new Node(data); if (data < root.data) root.left = insert(root.left, data); if (data > root.data) root.right = insert(root.right, data); return root; } // InOrder function to display value of array // in sorted order function inOrder(root) { if (root == null) return; inOrder(root.left); document.write(root.data + " "); inOrder(root.right); } // Driver Code var arr = [1, 2, 3, 2, 5, 4, 4 ]; // Finding size of array arr[] var n = arr.length; var root = null; for(var i = 0; i < n; i++) { // Insert element of arr[] in BST root = insert(root, arr[i]); } // Inorder Traversal to print nodes of Tree inOrder(root); </script>
1 2 3 4 5
Complejidad de tiempo: en el peor de los casos (cuando se ordena la array) donde N es el tamaño de la array dada.
Espacio Auxiliar : O(N).