Duplicates Removal in Array using BST

Given an array arr[] of integers, the task is to remove duplicates from the given array.

Examples:

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}

The duplicates in the array can be removed using Binary Search Tree. The idea is to create a Binary Seearch Tree using the array elements with the condition that the first element is taken as the root(parent) element and when the element “less” than root appears, it is made the left child and the element “greater” than root is made the right child of the root. Since no condition for “equal” exists the duplicates are automatically removed when we form binary search tree from the array elements.



For the array, arr[] = {1, 2, 3, 2, 5, 4, 4}

BST will be:

Approach:

  • Form BST using the array elements
  • Display the elements using any Tree Traversal method.

Below is the implementation of the above approach.

C

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// 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;
}

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Python3

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# Python3 implmentation 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

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Output:

1 2 3 4 5

Time Complexity: O(N) where N is the size of given array.



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Improved By : AnkitRai01