Find k-th smallest element in BST (Order Statistics in BST)

Given the root of a binary search tree and K as input, find Kth smallest element in BST. 
For example, in the following BST, if k = 3, then the output should be 10, and if k = 5, then the output should be 14.

Method 1: Using Inorder Traversal (O(n) time and O(h) auxiliary space) 

The Inorder Traversal of a BST traverses the nodes in increasing order. So the idea is to traverse the tree in Inorder. While traversing, keep track of the count of the nodes visited. If the count becomes k, print the node. 

C++

// A simple inorder traversal based C++ program to find k-th
// smallest element in a BST.
#include <iostream>

using namespace std;

// A BST node
struct Node {
    int data;
    Node *left, *right;
    Node(int x)
    {
        data = x;
        left = right = NULL;
    }
};

// Recursive function to insert an key into BST
Node* insert(Node* root, int x)
{
    if (root == NULL)
        return new Node(x);
    if (x < root->data)
        root->left = insert(root->left, x);
    else if (x > root->data)
        root->right = insert(root->right, x);
    return root;
}

// Function to find k'th smallest element in BST
// Here count denotes the number of nodes processed so far
int count = 0;
Node* kthSmallest(Node* root, int& k)
{
    // base case
    if (root == NULL)
        return NULL;

    // search in left subtree
    Node* left = kthSmallest(root->left, k);

    // if k'th smallest is found in left subtree, return it
    if (left != NULL)
        return left;

    // if current element is k'th smallest, return it
    count++;
    if (count == k)
        return root;

    // else search in right subtree
    return kthSmallest(root->right, k);
}

// Function to print k'th smallest element in BST
void printKthSmallest(Node* root, int k)
{
    // maintain index to count number of nodes processed so far

    Node* res = kthSmallest(root, k);
    if (res == NULL)
        cout << "There are less than k nodes in the BST";
    else
        cout << "K-th Smallest Element is " << res->data;
}

// main function
int main()
{
    Node* root = NULL;
    int keys[] = { 20, 8, 22, 4, 12, 10, 14 };

    for (int x : keys)
        root = insert(root, x);

    int k = 3;
    printKthSmallest(root, k);
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

C

// A simple inorder traversal based C++ program to find k-th
// smallest element in a BST.
#include <stdio.h>
#include <stdlib.h>

// A BST node
typedef struct Node {
    int data;
    struct Node *left, *right;
} Node;

struct Node* new_node(int x)
{
    struct Node* p = malloc(sizeof(struct Node));
    p->data = x;
    p->left = NULL;
    p->right = NULL;
    return p;
}

// Recursive function to insert an key into BST
Node* insert(Node* root, int x)
{
    if (root == NULL)
        return new_node(x);
    if (x < root->data)
        root->left = insert(root->left, x);
    else if (x > root->data)
        root->right = insert(root->right, x);
    return root;
}

// Function to find k'th smallest element in BST
// Here count denotes the number of nodes processed so far
int count = 0;
Node* kthSmallest(Node* root, int k)
{
    // base case
    if (root == NULL)
        return NULL;

    // search in left subtree
    Node* left = kthSmallest(root->left, k);

    // if k'th smallest is found in left subtree, return it
    if (left != NULL)
        return left;

    // if current element is k'th smallest, return it
    count++;
    if (count == k)
        return root;

    // else search in right subtree
    return kthSmallest(root->right, k);
}

// Function to print k'th smallest element in BST
void printKthSmallest(Node* root, int k)
{
    // maintain index to count number of nodes processed so far
    Node* res = kthSmallest(root, k);
    if (res == NULL)
        printf("There are less than k nodes in the BST");
    else
        printf("K-th Smallest Element is %d", res->data);
}

// main function
int main()
{
    Node* root = NULL;
    int keys[] = { 20, 8, 22, 4, 12, 10, 14 };
    int keys_size = sizeof(keys) / sizeof(keys[0]);

    for (int i = 0; i < keys_size; i++)
        root = insert(root, keys[i]);

    int k = 3;
    printKthSmallest(root, k);
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// A simple inorder traversal based Java program
// to find k-th smallest element in a BST.

import java.io.*;
// A BST node
class Node {
    int data;
    Node left, right;
    Node(int x)
    {
        data = x;
        left = right = null;
    }
}

class GFG {

    static int count = 0;
    // Recursive function to insert an key into BST
    public static Node insert(Node root, int x)
    {
        if (root == null)
            return new Node(x);
        if (x < root.data)
            root.left = insert(root.left, x);
        else if (x > root.data)
            root.right = insert(root.right, x);
        return root;
    }

    // Function to find k'th smallest element in BST
    // Here count denotes the number of nodes processed so far
    public static Node kthSmallest(Node root, int k)
    {
        // base case
        if (root == null)
            return null;

        // search in left subtree
        Node left = kthSmallest(root.left, k);

        // if k'th smallest is found in left subtree, return it
        if (left != null)
            return left;

        // if current element is k'th smallest, return it
        count++;
        if (count == k)
            return root;

        // else search in right subtree
        return kthSmallest(root.right, k);
    }

    // Function to find k'th smallest element in BST
    public static void printKthSmallest(Node root, int k)
    {
        Node res = kthSmallest(root, k);
        if (res == null)
            System.out.println("There are less than k nodes in the BST");
        else
            System.out.println("K-th Smallest Element is " + res.data);
    }

    public static void main(String[] args)
    {
        Node root = null;
        int keys[] = { 20, 8, 22, 4, 12, 10, 14 };
        for (int x : keys)
            root = insert(root, x);
        int k = 3;
        printKthSmallest(root, k);
    }
}

// This code is contributed by Aditya Kumar (adityakumar129)

Python3

# A simple inorder traversal based Python3
# program to find k-th smallest element
# in a BST.

# A BST node


class Node:

    def __init__(self, key):

        self.data = key
        self.left = None
        self.right = None

# Recursive function to insert an key into BST


def insert(root, x):

    if (root == None):
        return Node(x)
    if (x < root.data):
        root.left = insert(root.left, x)
    elif (x > root.data):
        root.right = insert(root.right, x)
    return root

# Function to find k'th largest element
# in BST. Here count denotes the number
# of nodes processed so far


def kthSmallest(root):

    global k

    # Base case
    if (root == None):
        return None

    # Search in left subtree
    left = kthSmallest(root.left)

    # If k'th smallest is found in
    # left subtree, return it
    if (left != None):
        return left

    # If current element is k'th
    # smallest, return it
    k -= 1
    if (k == 0):
        return root

    # Else search in right subtree
    return kthSmallest(root.right)

# Function to find k'th largest element in BST


def printKthSmallest(root):

    res = kthSmallest(root)

    if (res == None):
        print("There are less than k nodes in the BST")
    else:
        print("K-th Smallest Element is ", res.data)


# Driver code
if __name__ == '__main__':

    root = None
    keys = [20, 8, 22, 4, 12, 10, 14]

    for x in keys:
        root = insert(root, x)

    k = 3

    printKthSmallest(root)

# This code is contributed by mohit kumar 29

C#

// A simple inorder traversal
// based C# program to find
// k-th smallest element in a BST.
using System;

// A BST node
class Node{

public int data;
public Node left, right;
public Node(int x)
{
  data = x;
  left = right = null;
}
}

class GFG{

static int count = 0;

// Recursive function to
// insert an key into BST
public static Node insert(Node root,
                          int x)
{
  if (root == null)
    return new Node(x);
  if (x < root.data)
    root.left = insert(root.left, x);
  else if (x > root.data)
    root.right = insert(root.right, x);
  return root;
}

// Function to find k'th largest
// element in BST. Here count
// denotes the number of nodes
// processed so far
public static Node kthSmallest(Node root,
                               int k)
{
  // base case
  if (root == null)
    return null;

  // search in left subtree
  Node left = kthSmallest(root.left, k);

  // if k'th smallest is found
  // in left subtree, return it
  if (left != null)
    return left;

  // if current element is
  // k'th smallest, return it
  count++;
  if (count == k)
    return root;

  // else search in right subtree
  return kthSmallest(root.right, k);
}

// Function to find k'th largest
// element in BST
public static void printKthSmallest(Node root,
                                    int k)
{
  // Maintain an index to
  // count number of nodes
  // processed so far
  count = 0;

  Node res = kthSmallest(root, k);

  if (res == null)
    Console.WriteLine("There are less " +
                      "than k nodes in the BST");
  else
    Console.WriteLine("K-th Smallest" +
                      " Element is " + res.data);
}

// Driver code
public static void Main(String[] args)
{

  Node root = null;
  int []keys = {20, 8, 22, 4,
                12, 10, 14};

  foreach (int x in keys)
    root = insert(root, x);

  int k = 3;
  printKthSmallest(root, k);
}
}

// This code is contributed by gauravrajput1

Javascript

<script>
// A simple inorder traversal based Javascript program
// to find k-th smallest element in a BST.

    // A BST node
    class Node
    {
        constructor(x) {
            this.data = x;
            this.left = null;
            this.right = null;
          }
    }

    let count = 0;

    // Recursive function to insert an key into BST
    function insert(root,x)
    {
        if (root == null)
            return new Node(x);
        if (x < root.data)
            root.left = insert(root.left, x);
        else if (x > root.data)
            root.right = insert(root.right, x);
        return root;
    }

    // Function to find k'th largest element in BST
    // Here count denotes the number
    // of nodes processed so far
    function kthSmallest(root,k)
    {
        // base case
        if (root == null)
            return null;

        // search in left subtree
        let left = kthSmallest(root.left, k);

        // if k'th smallest is found in left subtree, return it
        if (left != null)
            return left;

        // if current element is k'th smallest, return it
        count++;
        if (count == k)
            return root;

        // else search in right subtree
        return kthSmallest(root.right, k);
    }

    // Function to find k'th largest element in BST
    function printKthSmallest(root,k)
    {
        // maintain an index to count number of
        // nodes processed so far
        count = 0;

        let res = kthSmallest(root, k);
        if (res == null)
            document.write("There are less "
                        + "than k nodes in the BST");
        else
            document.write("K-th Smallest"
                    + " Element is " + res.data);
    }

    let root=null;
    let key=[20, 8, 22, 4, 12, 10, 14 ];
    for(let i=0;i<key.length;i++)
    {
        root = insert(root, key[i]);
    }

    let k = 3;
    printKthSmallest(root, k);



    // This code is contributed by unknown2108
</script>

Output

K-th Smallest Element is 10

Time complexity: O(h) where h is the height of the tree.
Auxiliary Space: O(h)

We can optimize space using Morris Traversal. Please refer K’th smallest element in BST using O(1) Extra Space for details.

Method 2: Using Any Tree Traversal (pre-in-post) than return kth smallest easily.

Approach:  

         Here we use pre order traversal than sort it and return the kth smallest element.

 Algorithm:
    Here we have tree we will take preorder of it as :
    

preorder will : 20 8 4 12 10 14 22

  And store it in array/vector.
  After taking preorder we will sort it and than return k-1 element from the array.

C++

// A simple inorder traversal based C++ program to find k-th
// smallest element in a BST.
#include <iostream>
#include <bits/stdc++.h>
using namespace std;

// A BST node
struct Node {
    int data;
    Node *left, *right;
    int lCount;
    Node(int x)
    {
        data = x;
        left = right = NULL;
        lCount = 0;
    }
};

// Recursive function to insert an key into BST
Node* insert(Node* root, int x)
{
    if (root == NULL)
        return new Node(x);

    // If a node is inserted in left subtree, then lCount of
    // this node is increased. For simplicity, we are
    // assuming that all keys (tried to be inserted) are
    // distinct.
    if (x < root->data) {
        root->left = insert(root->left, x);
        root->lCount++;
    }

    else if (x > root->data)
        root->right = insert(root->right, x);
    return root;
}
//preorder function.
void preorder(Node* root,vector<int>&v){
    if(root==NULL)return;
    v.push_back(root->data);
    preorder(root->left,v);
    preorder(root->right,v);
}
int main(){
    Node* root = NULL;
    int keys[] = { 20, 8, 22, 4, 12, 10, 14 };
    for (int x : keys)
        root = insert(root, x);
    int k = 4;
    vector<int>v;
    preorder(root,v);
    //for(auto it:v)cout<<it<<" ";
   // cout<<endl;
  //sorting the given vector.
    sort(v.begin(),v.end());
   //return kth smallest element.
    cout<<v[k-1]<<endl;
  //code and approach contributed by Sanket Gode.
    return 0;
}

Java

import java.util.*;

class Node {
    int data;
    Node left, right;
    int lCount;
    Node(int x) {
        data = x;
        left = right = null;
        lCount = 0;
    }
}

public class KthSmallestElementBST {
    // Recursive function to insert a key into BST
    static Node insert(Node root, int x) {
        if (root == null)
            return new Node(x);

        // If a node is inserted in left subtree, then lCount of
        // this node is increased. For simplicity, we are
        // assuming that all keys (tried to be inserted) are
        // distinct.
        if (x < root.data) {
            root.left = insert(root.left, x);
            root.lCount++;
        }
        else if (x > root.data)
            root.right = insert(root.right, x);
        return root;
    }

    // Preorder traversal of BST
    static void preorder(Node root, List<Integer> v) {
        if (root == null) return;
        v.add(root.data);
        preorder(root.left, v);
        preorder(root.right, v);
    }

    public static void main(String[] args) {
        Node root = null;
        int[] keys = { 20, 8, 22, 4, 12, 10, 14 };
        for (int x : keys)
            root = insert(root, x);
        int k = 4;
        List<Integer> v = new ArrayList<Integer>();
        preorder(root, v);

        // Sorting the given vector
        Collections.sort(v);

        // Finding kth smallest element
        System.out.println(v.get(k-1));
    }
}

Python3

# A BST node
class Node:
    def __init__(self, x):
        self.data = x
        self.left = None
        self.right = None
        self.lCount = 0

# Recursive function to insert a key into BST
def insert(root, x):
    if root is None:
        return Node(x)

    # If a node is inserted in the left subtree, then lCount of
    # this node is increased. For simplicity, we are
    # assuming that all keys (tried to be inserted) are
    # distinct.
    if x < root.data:
        root.left = insert(root.left, x)
        root.lCount += 1
    elif x > root.data:
        root.right = insert(root.right, x)
    return root

# Preorder traversal function
def preorder(root, v):
    if root is None:
        return
    v.append(root.data)
    preorder(root.left, v)
    preorder(root.right, v)

# Main function
if __name__ == '__main__':
    root = None
    keys = [20, 8, 22, 4, 12, 10, 14]
    for x in keys:
        root = insert(root, x)
    k = 4
    v = []
    preorder(root, v)

    # Sorting the given list
    v.sort()

    # Return the kth smallest element
    print(v[k-1])

C#

using System;
using System.Collections.Generic;

class Program
{
    // A BST node
    class Node
    {
        public int data;
        public Node left, right;
        public int lCount;
        public Node(int x)
        {
            data = x;
            left = right = null;
            lCount = 0;
        }
    }

    // Recursive function to insert an key into BST
    static Node insert(Node root, int x)
    {
        if (root == null)
            return new Node(x);

        // If a node is inserted in left subtree, then lCount of
        // this node is increased. For simplicity, we are
        // assuming that all keys (tried to be inserted) are
        // distinct.
        if (x < root.data)
        {
            root.left = insert(root.left, x);
            root.lCount++;
        }
        else if (x > root.data)
        {
            root.right = insert(root.right, x);
        }
        return root;
    }

    // Inorder traversal to get sorted order
    static void inorder(Node root, List<int> v)
    {
        if (root == null) return;
        inorder(root.left, v);
        v.Add(root.data);
        inorder(root.right, v);
    }

    static void Main(string[] args)
    {
        Node root = null;
        int[] keys = { 20, 8, 22, 4, 12, 10, 14 };
        foreach (int x in keys)
        {
            root = insert(root, x);
        }

        int k = 4;
        List<int> v = new List<int>();
        inorder(root, v);

        // k-th smallest element is at index k-1 in the sorted array
        Console.WriteLine(v[k - 1]);
    }
}

Javascript

class Node {
  constructor(x) {
    this.data = x;
    this.left = null;
    this.right = null;
    this.lCount = 0;
  }
}

function insert(root, x) {
  if (root === null) {
    return new Node(x);
  }

  if (x < root.data) {
    root.left = insert(root.left, x);
    root.lCount++;
  } else if (x > root.data) {
    root.right = insert(root.right, x);
  }
  return root;
}

function preorder(root, v) {
  if (root === null) {
    return;
  }

  v.push(root.data);
  preorder(root.left, v);
  preorder(root.right, v);
}

let root = null;
const keys = [20, 8, 22, 4, 12, 10, 14];

for (const x of keys) {
  root = insert(root, x);
}

const k = 4;
const v = [];

preorder(root, v);

v.sort((a, b) => a - b);

console.log(v[k - 1]);

Output

12

Complexity Analysis:

Time Complexity: O(nlogn) i.e O(n) time for preorder and nlogn time for sorting.
Auxiliary Space: O(n) i.e for vector/array storage.

Method 3: Augmented Tree Data Structure (O(h) Time Complexity and O(h) auxiliary space)

The idea is to maintain the rank of each node. We can keep track of elements in the left subtree of every node while building the tree. Since we need the K-th smallest element, we can maintain the number of elements of the left subtree in every node.
Assume that the root is having ‘lCount’ nodes in its left subtree. If K = lCount + 1, root is K-th node. If K < lCount + 1, we will continue our search (recursion) for the Kth smallest element in the left subtree of root. If K > lCount + 1, we continue our search in the right subtree for the (K – lCount – 1)-th smallest element. Note that we need the count of elements in the left subtree only.

C++

// A simple inorder traversal based C++ program to find k-th
// smallest element in a BST.
#include <iostream>
using namespace std;

// A BST node
struct Node {
    int data;
    Node *left, *right;
    int lCount;
    Node(int x)
    {
        data = x;
        left = right = NULL;
        lCount = 0;
    }
};

// Recursive function to insert an key into BST
Node* insert(Node* root, int x)
{
    if (root == NULL)
        return new Node(x);

    // If a node is inserted in left subtree, then lCount of
    // this node is increased. For simplicity, we are
    // assuming that all keys (tried to be inserted) are
    // distinct.
    if (x < root->data) {
        root->left = insert(root->left, x);
        root->lCount++;
    }

    else if (x > root->data)
        root->right = insert(root->right, x);
    return root;
}

// Function to find k'th smallest element in BST
// Here count denotes the number of nodes processed so far
Node* kthSmallest(Node* root, int k)
{
    // base case
    if (root == NULL)
        return NULL;
    int count = root->lCount + 1;
    if (count == k)
        return root;
    if (count > k)
        return kthSmallest(root->left, k);
    // else search in right subtree
    return kthSmallest(root->right, k - count);
}

// main function
int main()
{
    Node* root = NULL;
    int keys[] = { 20, 8, 22, 4, 12, 10, 14 };
    for (int x : keys)
        root = insert(root, x);
    int k = 4;
    Node* res = kthSmallest(root, k);
    if (res == NULL)
        cout << "There are less than k nodes in the BST";
    else
        cout << "K-th Smallest Element is " << res->data;
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

C

// A simple inorder traversal based C++ program to find k-th
// smallest element in a BST.
#include <stdio.h>
#include <stdlib.h>

// A BST node
typedef struct Node {
    int data;
    struct Node *left, *right;
    int lCount;
} Node;

Node* new_node(int x)
{
    Node* newNode = malloc(sizeof(Node));
    newNode->data = x;
    newNode->left = NULL;
    newNode->right = NULL;
    return newNode;
}

// Recursive function to insert an key into BST
Node* insert(Node* root, int x)
{
    if (root == NULL)
        return new_node(x);

    // If a node is inserted in left subtree, then lCount of
    // this node is increased. For simplicity, we are
    // assuming that all keys (tried to be inserted) are
    // distinct.
    if (x < root->data) {
        root->left = insert(root->left, x);
        root->lCount++;
    }

    else if (x > root->data)
        root->right = insert(root->right, x);
    return root;
}

// Function to find k'th smallest element in BST
// Here count denotes the number of nodes processed so far
Node* kthSmallest(Node* root, int k)
{
    // base case
    if (root == NULL)
        return NULL;
    int count = root->lCount + 1;
    if (count == k)
        return root;
    if (count > k)
        return kthSmallest(root->left, k);
    // else search in right subtree
    return kthSmallest(root->right, k - count);
}

// main function
int main()
{
    Node* root = NULL;
    int keys[] = { 20, 8, 22, 4, 12, 10, 14 };
    int keys_size = sizeof(keys) / sizeof(keys[0]);

    for (int i = 0; i < keys_size; i++)
        root = insert(root, keys[i]);
    int k = 4;
    Node* res = kthSmallest(root, k);
    if (res == NULL)
        printf("There are less than k nodes in the BST");
    else
        printf("K-th Smallest Element is %d", res->data);
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// A simple inorder traversal based Java program
// to find k-th smallest element in a BST.
import java.io.*;
import java.util.*;

// A BST node
class Node {
    int data;
    Node left, right;
    int lCount;
    Node(int x)
    {
        data = x;
        left = right = null;
        lCount = 0;
    }
}

class Gfg {
    // Recursive function to insert an key into BST
    public static Node insert(Node root, int x)
    {
        if (root == null)
            return new Node(x);

        // If a node is inserted in left subtree, then
        // lCount of this node is increased. For simplicity,
        // we are assuming that all keys (tried to be
        // inserted) are distinct.
        if (x < root.data) {
            root.left = insert(root.left, x);
            root.lCount++;
        }

        else if (x > root.data)
            root.right = insert(root.right, x);
        return root;
    }

    // Function to find k'th largest element in BST
    // Here count denotes the number of nodes processed so far
    public static Node kthSmallest(Node root, int k)
    {
        // base case
        if (root == null)
            return null;

        int count = root.lCount + 1;
        if (count == k)
            return root;

        if (count > k)
            return kthSmallest(root.left, k);

        // else search in right subtree
        return kthSmallest(root.right, k - count);
    }

    // main function
    public static void main(String args[])
    {
        Node root = null;
        int keys[] = { 20, 8, 22, 4, 12, 10, 14 };

        for (int x : keys)
            root = insert(root, x);

        int k = 4;
        Node res = kthSmallest(root, k);
        if (res == null)
            System.out.println("There are less than k nodes in the BST");
        else
            System.out.println("K-th Smallest Element is " + res.data);
    }
}

// This code is contributed by Aditya Kumar (adityakumar129)

Python3

# A simple inorder traversal based Python3
# program to find k-th smallest element in a BST.

# A BST node
class newNode:

    def __init__(self, x):

        self.data = x
        self.left = None
        self.right = None
        self.lCount = 0

# Recursive function to insert
# an key into BST
def insert(root, x):

    if (root == None):
        return newNode(x)

    # If a node is inserted in left subtree,
    # then lCount of this node is increased.
    # For simplicity, we are assuming that
    # all keys (tried to be inserted) are
    # distinct.
    if (x < root.data):
        root.left = insert(root.left, x)
        root.lCount += 1

    elif (x > root.data):
        root.right = insert(root.right, x);

    return root

# Function to find k'th largest element
# in BST. Here count denotes the number
# of nodes processed so far
def kthSmallest(root, k):

    # Base case
    if (root == None):
        return None

    count = root.lCount + 1

    if (count == k):
        return root

    if (count > k):
        return kthSmallest(root.left, k)

    # Else search in right subtree
    return kthSmallest(root.right, k - count)

# Driver code
if __name__ == '__main__':

    root = None
    keys = [ 20, 8, 22, 4, 12, 10, 14 ]

    for x in keys:
        root = insert(root, x)

    k = 4
    res = kthSmallest(root, k)

    if (res == None):
        print("There are less than k nodes in the BST")
    else:
        print("K-th Smallest Element is", res.data)

# This code is contributed by bgangwar59

C#

// A simple inorder traversal based C# program
// to find k-th smallest element in a BST.
using System;

// A BST node
public class Node
{
    public int data;
    public Node left, right;
    public int lCount;

    public Node(int x)
    {
        data = x;
        left = right = null;
        lCount = 0;
    }
}

class GFG{

// Recursive function to insert an key into BST
public static Node insert(Node root, int x)
{
    if (root == null)
        return new Node(x);

    // If a node is inserted in left subtree,
    // then lCount of this node is increased.
    // For simplicity, we are assuming that
    // all keys (tried to be inserted) are
    // distinct.
    if (x < root.data)
    {
        root.left = insert(root.left, x);
        root.lCount++;
    }

    else if (x > root.data)
        root.right = insert(root.right, x);

    return root;
}

// Function to find k'th largest element
// in BST. Here count denotes the number
// of nodes processed so far
public static Node kthSmallest(Node root, int k)
{

    // Base case
    if (root == null)
        return null;

    int count = root.lCount + 1;
    if (count == k)
        return root;

    if (count > k)
        return kthSmallest(root.left, k);

    // Else search in right subtree
    return kthSmallest(root.right, k - count);
}

// Driver Code
public static void Main(String[] args)
{
    Node root = null;
    int[] keys = { 20, 8, 22, 4, 12, 10, 14 };

    foreach(int x in keys)
        root = insert(root, x);

    int k = 4;
    Node res = kthSmallest(root, k);

    if (res == null)
        Console.WriteLine("There are less " +
                          "than k nodes in the BST");
    else
        Console.WriteLine("K-th Smallest" +
                          " Element is " + res.data);
}
}

// This code is contributed by aashish1995

Javascript

<script>

// A simple inorder traversal based
// Javascript program to find k-th
// smallest element in a BST.

// A BST node
class Node
{
    constructor(x)
    {
        this.data = x;
        this.left = null;
        this.right = null;
        this.lCount = 0;
    }
}

// Recursive function to insert an key into BST
function insert(root, x)
{
    if (root == null)
        return new Node(x);

    // If a node is inserted in left subtree,
    // then lCount of this node is increased.
    // For simplicity, we are assuming that
    // all keys (tried to be inserted) are
    // distinct.
    if (x < root.data)
    {
        root.left = insert(root.left, x);
        root.lCount++;
    }

    else if (x > root.data)
        root.right = insert(root.right, x);

    return root;
}

// Function to find k'th largest element
// in BST. Here count denotes the number
// of nodes processed so far
function kthSmallest(root, k)
{

    // Base case
    if (root == null)
        return null;

    let count = root.lCount + 1;
    if (count == k)
        return root;

    if (count > k)
        return kthSmallest(root.left, k);

    // Else search in right subtree
    return kthSmallest(root.right, k - count);
}

// Driver code
let root = null;
let keys = [ 20, 8, 22, 4, 12, 10, 14 ];

for(let x = 0; x < keys.length; x++)
    root = insert(root, keys[x]);

let k = 4;
let res = kthSmallest(root, k);

if (res == null)
    document.write("There are less than k " +
                   "nodes in the BST" + "</br>");
else
    document.write("K-th Smallest" +
                   " Element is " + res.data);

// This code is contributed by divyeshrabadiya07

</script>

Output

K-th Smallest Element is 12

Time complexity: O(h) where h is the height of the tree.
Auxiliary Space: O(h)

Method 4: Iterative approach using Stack: The basic idea behind the Iterative Approach using Stack to find the kth smallest element in a Binary Search Tree (BST) is to traverse the tree in an inorder fashion using a stack until we find the k^th smallest element. Follow the steps to implement the above idea:

1. Create an empty stack and set the current node to the root of the BST.
2. Push all the left subtree nodes of the current node onto the stack until the current node is NULL.
3. Pop the top node from the stack and check if it is the k-th element. If it is, return its value.
4. Decrement the value of k by 1.
5. Set the current node to the right child of the popped node.
6. Go to step 2 if the stack is not empty or k is not equal to 0.

Below is the implementation of the above approach:

C++

// C++ code to implement the iterative approach
#include <iostream>
#include <stack>
using namespace std;

// Definition of a BST node
struct Node {
    int data;
    Node *left, *right;
};

// Function to create a new BST node
Node* newNode(int key)
{
    Node* node = new Node;
    node->data = key;
    node->left = node->right = NULL;
    return node;
}

// Function to insert a new node in BST
Node* insert(Node* root, int key)
{
    // If the tree is empty, return a new node
    if (root == NULL)
        return newNode(key);

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

    // Return the (unchanged) node pointer
    return root;
}

// Function to find the k-th smallest
// element in BST
int kthSmallest(Node* root, int k)
{
    // Create an empty stack
    stack<Node*> s;

    // Loop until stack is empty or
    // k becomes zero
    while (root != NULL || !s.empty()) {

        // Push all the left subtree
        // nodes onto the stack
        while (root != NULL) {
            s.push(root);
            root = root->left;
        }

        // Pop the top node from the
        // stack and check if it is
        // the k-th element
        root = s.top();
        s.pop();

        if (--k == 0)
            return root->data;

        // Set root to the right child
        // and continue with the traversal
        root = root->right;
    }

    // If k is greater than the number
    // of nodes in BST, return -1
    return -1;
}

// Driver Code
int main()
{
    Node* root = NULL;
    int keys[] = { 20, 8, 22, 4, 12, 10, 14 };

    // Insert all the keys into BST
    for (int x : keys)
        root = insert(root, x);

    int k = 4;

    // Find the k-th smallest element in BST
    int kth_smallest = kthSmallest(root, k);
    if (kth_smallest != -1)
        cout << "K-th smallest element in BST is: "
             << kth_smallest << endl;
    else
        cout << "Invalid input" << endl;

    return 0;
}

Python3

# Python code to implement the iterative approach

# Definition of a BST node


class Node:
    def __init__(self, key):
        self.data = key
        self.left = None
        self.right = None

# Function to insert a new node in BST


def insert(root, key):
    # If the tree is empty, return a new node
    if root is None:
        return Node(key)

    # Otherwise, recur down the tree
    if key < root.data:
        root.left = insert(root.left, key)
    elif key > root.data:
        root.right = insert(root.right, key)

    # Return the (unchanged) node pointer
    return root

# Function to find the k-th smallest
# element in BST


def kthSmallest(root, k):
    # Create an empty stack
    stack = []

    # Loop until stack is empty or
    # k becomes zero
    while root is not None or len(stack) > 0:

        # Push all the left subtree
        # nodes onto the stack
        while root is not None:
            stack.append(root)
            root = root.left

        # Pop the top node from the
        # stack and check if it is
        # the k-th element
        root = stack.pop()
        k -= 1

        if k == 0:
            return root.data

        # Set root to the right child
        # and continue with the traversal
        root = root.right

    # If k is greater than the number
    # of nodes in BST, return -1
    return -1


# Driver Code
if __name__ == '__main__':
    root = None
    keys = [20, 8, 22, 4, 12, 10, 14]

    # Insert all the keys into BST
    for x in keys:
        root = insert(root, x)

    k = 4

    # Find the k-th smallest element in BST
    kth_smallest = kthSmallest(root, k)
    if kth_smallest != -1:
        print("K-th smallest element in BST is:", kth_smallest)
    else:
        print("Invalid input")

Output

K-th smallest element in BST is: 12

Time Complexity: O(h+ k), The time complexity of the Iterative Approach using Stack to find the kth smallest element in a BST is O(h + k), where h is the height of the BST and k is the value of k.

Auxiliary Space: O(h+k), The space complexity of the code is O(h + k), where h is the height of the BST and k is the maximum size of the stack.