Tree Sort

Tree sort is a sorting algorithm that is based on Binary Search Tree data structure. It first creates a binary search tree from the elements of the input list or array and then performs an in-order traversal on the created binary search tree to get the elements in sorted order.

Algorithm:

Step 1: Take the elements input in an array.
Step 2: Create a Binary search tree by inserting data items from the array into the
        binary search tree.
Step 3: Perform in-order traversal on the tree to get the elements in sorted order.



C++

// C++ program to implement Tree Sort
#include<bits/stdc++.h>

using namespace std;

struct Node
{
    int key;
    struct Node *left, *right;
};

// A utility function to create a new BST Node
struct Node *newNode(int item)
{
    struct Node *temp = new Node;
    temp->key = item;
    temp->left = temp->right = NULL;
    return temp;
}

// Stores inoder traversal of the BST
// in arr[]
void storeSorted(Node *root, int arr[], int &i)
{
    if (root != NULL)
    {
        storeSorted(root->left, arr, i);
        arr[i++] = root->key;
        storeSorted(root->right, arr, i);
    }
}

/* A utility function to insert a new
   Node with given key in BST */
Node* insert(Node* node, int key)
{
    /* If the tree is empty, return a new Node */
    if (node == NULL) return newNode(key);

    /* Otherwise, recur down the tree */
    if (key < node->key)
        node->left  = insert(node->left, key);
    else if (key > node->key)
        node->right = insert(node->right, key);

    /* return the (unchanged) Node pointer */
    return node;
}

// This function sorts arr[0..n-1] using Tree Sort
void treeSort(int arr[], int n)
{
    struct Node *root = NULL;

    // Construct the BST
    root = insert(root, arr[0]);
    for (int i=1; i<n; i++)
        insert(root, arr[i]);

    // Store inoder traversal of the BST
    // in arr[]
    int i = 0;
    storeSorted(root, arr, i);
}

// Driver Program to test above functions
int main()
{
    //create input array
    int arr[] = {5, 4, 7, 2, 11};
    int n = sizeof(arr)/sizeof(arr[0]);

    treeSort(arr, n);

        for (int i=0; i<n; i++)
       cout << arr[i] << " ";

    return 0;
}

Java

// Java program to 
// implement Tree Sort
class GFG 
{

    // Class containing left and
    // right child of current 
    // node and key value
    class Node 
    {
        int key;
        Node left, right;

        public Node(int item) 
        {
            key = item;
            left = right = null;
        }
    }

    // Root of BST
    Node root;

    // Constructor
    GFG() 
    { 
        root = null; 
    }

    // This method mainly
    // calls insertRec()
    void insert(int key)
    {
        root = insertRec(root, key);
    }
    
    /* A recursive function to 
    insert a new key in BST */
    Node insertRec(Node root, int key) 
    {

        /* If the tree is empty,
        return a new node */
        if (root == null) 
        {
            root = new Node(key);
            return root;
        }

        /* Otherwise, recur
        down the tree */
        if (key < root.key)
            root.left = insertRec(root.left, key);
        else if (key > root.key)
            root.right = insertRec(root.right, key);

        /* return the root */
        return root;
    }
    
    // A function to do 
    // inorder traversal of BST
    void inorderRec(Node root) 
    {
        if (root != null) 
        {
            inorderRec(root.left);
            System.out.print(root.key + " ");
            inorderRec(root.right);
        }
    }
    void treeins(int arr[])
    {
        for(int i = 0; i < arr.length; i++)
        {
            insert(arr[i]);
        }
        
    }

    // Driver Code
    public static void main(String[] args) 
    {
        GFG tree = new GFG();
        int arr[] = {5, 4, 7, 2, 11};
        tree.treeins(arr);
        tree.inorderRec(tree.root);
    }
}

// This code is contributed
// by Vibin M



Output:

2 4 5 7 11

Average Case Time Complexity : O(n log n) Adding one item to a Binary Search tree on average takes O(log n) time. Therefore, adding n items will take O(n log n) time

Worst Case Time Complexity : O(n2). The worst case time complexity of Tree Sort can be improved by using a self-balancing binary search tree like Red Black Tree, AVL Tree. Using self-balancing binary tree Tree Sort will take O(n log n) time to sort the array in worst case.

Auxiliary Space : O(n)

References:
https://en.wikipedia.org/wiki/Tree_sort

This article is contributed by Harsh Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : vibi, vsushko




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