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 nodestruct Node { int data; Node *left, *right; Node(int x) { data = x; left = right = NULL; }};// Recursive function to insert an key into BSTNode* 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 farint 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 BSTvoid 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 functionint 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 nodetypedef 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 BSTNode* 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 farint 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 BSTvoid 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 functionint 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 nodeclass 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 nodeclass Node: def __init__(self, key): self.data = key self.left = None self.right = None# Recursive function to insert an key into BSTdef 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 fardef 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 BSTdef 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 codeif __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 nodeclass 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 BSTpublic 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 farpublic 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 BSTpublic 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 codepublic 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(n) where n is the number of nodes in a binary search tree.
Auxiliary Space: O(h) where h is the height of the binary search tree.
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 nodestruct 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 BSTNode* 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 nodeclass Node: def __init__(self, x): self.data = x self.left = None self.right = None self.lCount = 0# Recursive function to insert a key into BSTdef 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 functiondef preorder(root, v): if root is None: return v.append(root.data) preorder(root.left, v) preorder(root.right, v)# Main functionif __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 nodestruct 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 BSTNode* 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 farNode* 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 functionint 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 nodetypedef 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 BSTNode* 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 farNode* 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 functionint 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 nodeclass 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 nodeclass newNode: def __init__(self, x): self.data = x self.left = None self.right = None self.lCount = 0# Recursive function to insert# an key into BSTdef 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 fardef 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 codeif __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 nodepublic 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 BSTpublic 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 farpublic 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 Codepublic 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 nodeclass Node{ constructor(x) { this.data = x; this.left = null; this.right = null; this.lCount = 0; }}// Recursive function to insert an key into BSTfunction 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 farfunction 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 codelet 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 kth smallest element. Follow the steps to implement the above idea:
- Create an empty stack and set the current node to the root of the BST.
- Push all the left subtree nodes of the current node onto the stack until the current node is NULL.
- Pop the top node from the stack and check if it is the k-th element. If it is, return its value.
- Decrement the value of k by 1.
- Set the current node to the right child of the popped node.
- 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 nodestruct Node { int data; Node *left, *right;};// Function to create a new BST nodeNode* newNode(int key){ Node* node = new Node; node->data = key; node->left = node->right = NULL; return node;}// Function to insert a new node in BSTNode* 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 BSTint 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 Codeint 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;} |
Java
import java.util.Stack;class Node { int data; Node left, right; Node(int key) { data = key; left = right = null; }}class GFG { static Node insert(Node root, int key) { if (root == null) return new Node(key); if (key < root.data) root.left = insert(root.left, key); else if (key > root.data) root.right = insert(root.right, key); return root; } static int kthSmallest(Node root, int k) { Stack<Node> stack = new Stack<>(); while (root != null || !stack.empty()) { while (root != null) { stack.push(root); root = root.left; } root = stack.pop(); if (--k == 0) return root.data; root = root.right; } return -1; } 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; int kth_smallest = kthSmallest(root, k); if (kth_smallest != -1) System.out.println("K-th smallest element in BST is: " + kth_smallest); else System.out.println("Invalid input"); }} |
Python3
# Python code to implement the iterative approach# Definition of a BST nodeclass Node: def __init__(self, key): self.data = key self.left = None self.right = None# Function to insert a new node in BSTdef 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 BSTdef 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 Codeif __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") |
C#
using System;using System.Collections.Generic;// Definition of a BST nodepublic class Node{ public int data; public Node left, right; public Node(int key) { data = key; left = right = null; }}public class BinarySearchTree{ // Function to insert a new node in BST public static Node Insert(Node root, int key) { // If the tree is empty, return a new node if (root == null) return new Node(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 public static int KthSmallest(Node root, int k) { // Create an empty stack Stack<Node> stack = new Stack<Node>(); // Loop until the stack is empty or k becomes zero while (root != null || stack.Count > 0) { // Push all the left subtree nodes onto the stack while (root != null) { stack.Push(root); root = root.left; } // Pop the top node from the stack and check if it is the k-th element root = stack.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 public static void Main() { Node root = null; int[] keys = { 20, 8, 22, 4, 12, 10, 14 }; // Insert all the keys into BST foreach (int key in keys) root = Insert(root, key); int k = 4; // Find the k-th smallest element in BST int kth_smallest = KthSmallest(root, k); if (kth_smallest != -1) Console.WriteLine("K-th smallest element in BST is: " + kth_smallest); else Console.WriteLine("Invalid input"); }} |
Javascript
// Definition of a BST nodeclass Node { constructor(data) { this.data = data; this.left = null; this.right = null; }}// Function to insert a new node in BSTfunction insert(root, key) { // If the tree is empty, return a new node if (root === null) return new Node(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 BSTfunction kthSmallest(root, k) { // Create an empty stack const stack = []; // Loop until stack is empty or k becomes zero while (root !== null || stack.length > 0) { // Push all the left subtree nodes onto the stack while (root !== null) { stack.push(root); root = root.left; } // Pop the top node from the stack and check if it is the k-th element root = stack.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 Codelet root = null;const keys = [20, 8, 22, 4, 12, 10, 14];// Insert all the keys into BSTfor (const x of keys) root = insert(root, x);const k = 4;// Find the k-th smallest element in BSTconst kth_smallest = kthSmallest(root, k);if (kth_smallest !== -1) console.log("K-th smallest element in BST is: " + kth_smallest);else console.log("Invalid input");// This code is contributed by Veerendra_Singh_Rajpoot |
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.
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