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Find Median in BST Subtree from Kth Smallest Node

Given a binary search tree, the task is to find the median of the subtree rooted with Kth smallest node of the tree.

Examples:

Input: N = 12, K = 9

50
/ \
25 75
/ \ \
15 45 90
/ \ / \ / \
11 20 30 49 80 100

Output: 85
Explanation: K=9. So the 9th smallest number is 75. The values of this subtree are 75 80 90 100. Now, the median of this is (80+90)/2 = 85

Approach: This can be solved with the following idea:

As it is BST, Inorder traversal of tree will give us sorted array. From there we can get the Kth smallest node. After that, iterate for that particular node and get the median.

Below are the steps involved:

  • Intialize a nodes of vector v.
  • Traverse in inorder, So as to get sorted vector.
  • Get the Kth smallest node.
  • Again Iterate in inorder traversal for kth samllest node and store them in another vector.
  • Once done, check the size of vector:
    • If even, median will be sum of both mid divide by 2.
    • If odd, median will be mid of vector.

Below is the implementation of the code:

C++

// Initial Template for C++
 
#include <bits/stdc++.h>
using namespace std;
 
// Creating a node structure
struct Node {
    int data;
    Node* left;
    Node* right;
};
 
// Function to create a new node
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
 
// Function to insert a new node in BST
Node* insert(Node* root, int data)
{
    if (root == NULL)
        root = newNode(data);
    else if (data < root->data)
        root->left = insert(root->left, data);
    else
        root->right = insert(root->right, data);
 
    return root;
}
 
// Function to iterate in inorder behaviour
void inorder(vector<Node*>& v, Node* root)
{
    if (root == NULL) {
        return;
    }
 
    inorder(v, root->left);
    v.push_back(root);
    inorder(v, root->right);
    return;
}
 
// Function to find median for Kth smallest node in BST
int median(Node* node, int k)
{
 
    // Intialization of vector of nodes
    vector<Node*> v;
 
    inorder(v, node);
 
    vector<Node*> v1;
 
    // Forming a vector for kth smallest node
    inorder(v1, v[k - 1]);
 
    // If size is even
    if (v1.size() % 2 == 0) {
        int mid = v1.size() / 2;
        return (v1[mid]->data + v1[mid - 1]->data) / 2;
    }
    return v1[v1.size() / 2]->data;
    return 0;
}
 
// Driver code
int main()
{
    Node* root = NULL;
    int n = 4;
    int k = 2;
 
    vector<int> nodes = { 3, 2, 4, 1 };
 
    for (int i = 0; i < n; i++) {
        int x = nodes[i];
        root = insert(root, x);
    }
 
    // Function call
    cout << median(root, k) << endl;
    return 0;
}

Java

/*code by Flutterfly*/
import java.util.ArrayList;
import java.util.List;
 
// Creating a node structure
class Node {
    int data;
    Node left, right;
 
    Node(int data) {
        this.data = data;
        left = right = null;
    }
}
 
public class Main {
 
    // Function to insert a new node in BST
    static Node insert(Node root, int data) {
        if (root == null)
            return new Node(data);
        else if (data < root.data)
            root.left = insert(root.left, data);
        else
            root.right = insert(root.right, data);
 
        return root;
    }
 
    // Function to iterate in inorder behavior
    static void inorder(List<Node> v, Node root) {
        if (root == null) {
            return;
        }
 
        inorder(v, root.left);
        v.add(root);
        inorder(v, root.right);
    }
 
    // Function to find median for Kth smallest node in BST
    static int median(Node node, int k) {
        List<Node> v = new ArrayList<>();
        inorder(v, node);
 
        List<Node> v1 = new ArrayList<>();
        inorder(v1, v.get(k - 1));
 
        // If size is even
        if (v1.size() % 2 == 0) {
            int mid = v1.size() / 2;
            return (v1.get(mid).data + v1.get(mid - 1).data) / 2;
        }
        return v1.get(v1.size() / 2).data;
    }
 
    // Driver code
    public static void main(String[] args) {
        Node root = null;
        int n = 4;
        int k = 2;
 
        int[] nodes = { 3, 2, 4, 1 };
 
        for (int i = 0; i < n; i++) {
            int x = nodes[i];
            root = insert(root, x);
        }
 
        // Function call
        System.out.println(median(root, k));
    }
}

Python3

# Creating a node class
class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
 
# Function to insert a new node in BST
def insert(root, data):
    if root is None:
        return Node(data)
    elif data < root.data:
        root.left = insert(root.left, data)
    else:
        root.right = insert(root.right, data)
 
    return root
 
# Function to iterate in inorder behavior
def inorder(v, root):
    if root is None:
        return
 
    inorder(v, root.left)
    v.append(root)
    inorder(v, root.right)
 
# Function to find median for Kth smallest node in BST
def median(node, k):
    # Initialization of a list of nodes
    v = []
 
    inorder(v, node)
 
    v1 = []
 
    # Forming a list for kth smallest node
    inorder(v1, v[k - 1])
 
    # If the size is even
    if len(v1) % 2 == 0:
        mid = len(v1) // 2
        return (v1[mid].data + v1[mid - 1].data) // 2
    return v1[len(v1) // 2].data
 
# Driver code
if __name__ == "__main__":
    root = None
    n = 4
    k = 2
 
    nodes = [3, 2, 4, 1]
 
    for x in nodes:
        root = insert(root, x)
 
    # Function call
    print(median(root, k))

C#

// code by Flutterfly
using System;
using System.Collections.Generic;
 
// Creating a node structure
public class Node
{
    public int data;
    public Node left, right;
 
    public Node(int data)
    {
        this.data = data;
        left = right = null;
    }
}
 
public class Program
{
    // Function to insert a new node in BST
    static Node Insert(Node root, int data)
    {
        if (root == null)
            return new Node(data);
        else if (data < root.data)
            root.left = Insert(root.left, data);
        else
            root.right = Insert(root.right, data);
 
        return root;
    }
 
    // Function to iterate in inorder behavior
    static void Inorder(List<Node> v, Node root)
    {
        if (root == null)
        {
            return;
        }
 
        Inorder(v, root.left);
        v.Add(root);
        Inorder(v, root.right);
    }
 
    // Function to find median for Kth smallest node in BST
    static int Median(Node node, int k)
    {
        List<Node> v = new List<Node>();
        Inorder(v, node);
 
        List<Node> v1 = new List<Node>();
        Inorder(v1, v[k - 1]);
 
        // If size is even
        if (v1.Count % 2 == 0)
        {
            int mid = v1.Count / 2;
            return (v1[mid].data + v1[mid - 1].data) / 2;
        }
        return v1[v1.Count / 2].data;
    }
 
    // Driver code
    public static void Main()
    {
        Node root = null;
        int n = 4;
        int k = 2;
 
        int[] nodes = { 3, 2, 4, 1 };
 
        for (int i = 0; i < n; i++)
        {
            int x = nodes[i];
            root = Insert(root, x);
        }
 
        // Function call
        Console.WriteLine(Median(root, k));
    }
}

Javascript

// Creating a node structure
class Node {
    constructor(data) {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
// Function to insert a new node in BST
function insert(root, data) {
    if (root === null)
        return new Node(data);
    else if (data < root.data)
        root.left = insert(root.left, data);
    else
        root.right = insert(root.right, data);
 
    return root;
}
 
// Function to iterate in inorder behavior
function inorder(v, root) {
    if (root === null) {
        return;
    }
 
    inorder(v, root.left);
    v.push(root);
    inorder(v, root.right);
}
 
// Function to find median for Kth smallest node in BST
function median(node, k) {
    const v = [];
    inorder(v, node);
 
    const v1 = [];
    inorder(v1, v[k - 1]);
 
    // If size is even
    if (v1.length % 2 === 0) {
        const mid = v1.length / 2;
        return Math.floor((v1[mid].data + v1[mid - 1].data) / 2);
    }
    return v1[Math.floor(v1.length / 2)].data;
}
 
// Driver code
let root = null;
const n = 4;
const k = 2;
 
const nodes = [3, 2, 4, 1];
 
for (let i = 0; i < n; i++) {
    const x = nodes[i];
    root = insert(root, x);
}
 
// Function call
console.log(median(root, k));

Output

1

Time Complexity: O(N)
Auxiliary Space: O(N)


 


Reffered: https://www.geeksforgeeks.org

DSA

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Category:
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Sub Category:
 Tutorial
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