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K’th Largest Element in BST when modification to BST is not allowed

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Given a Binary Search Tree (BST) and a positive integer k, find the k’th largest element in the Binary Search Tree. 
For example, in the following BST, if k = 3, then output should be 14, and if k = 5, then output should be 10. 

We have discussed two methods in this post. The method 1 requires O(n) time. The method 2 takes O(h) time where h is height of BST, but requires augmenting the BST (storing count of nodes in left subtree with every node). 
Can we find k’th largest element in better than O(n) time and no augmentation?

Approach: 

  1. The idea is to do reverse inorder traversal of BST. Keep a count of nodes visited.
  2. The reverse inorder traversal traverses all nodes in decreasing order, i.e, visit the right node then centre then left and continue traversing the nodes recursively.
  3. While doing the traversal, keep track of the count of nodes visited so far.
  4. When the count becomes equal to k, stop the traversal and print the key.

Implementation:

C++

// C++ program to find k'th largest element in BST
#include<bits/stdc++.h>
using namespace std;
 
struct Node
{
    int key;
    Node *left, *right;
};
 
// A utility function to create a new BST node
Node *newNode(int item)
{
    Node *temp = new Node;
    temp->key = item;
    temp->left = temp->right = NULL;
    return temp;
}
 
// A function to find k'th largest element in a given tree.
void kthLargestUtil(Node *root, int k, int &c)
{
    // Base cases, the second condition is important to
    // avoid unnecessary recursive calls
    if (root == NULL || c >= k)
        return;
 
    // Follow reverse inorder traversal so that the
    // largest element is visited first
    kthLargestUtil(root->right, k, c);
 
    // Increment count of visited nodes
    c++;
 
    // If c becomes k now, then this is the k'th largest
    if (c == k)
    {
        cout << "K'th largest element is "
             << root->key << endl;
        return;
    }
 
    // Recur for left subtree
    kthLargestUtil(root->left, k, c);
}
 
// Function to find k'th largest element
void kthLargest(Node *root, int k)
{
    // Initialize count of nodes visited as 0
    int c = 0;
 
    // Note that c is passed by reference
    kthLargestUtil(root, k, c);
}
 
/* 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;
}
 
// Driver Program to test above functions
int main()
{
    /* Let us create following BST
              50
           /     \
          30      70
         /  \    /  \
       20   40  60   80 */
    Node *root = NULL;
    root = insert(root, 50);
    insert(root, 30);
    insert(root, 20);
    insert(root, 40);
    insert(root, 70);
    insert(root, 60);
    insert(root, 80);
 
    int c = 0;
    for (int k=1; k<=7; k++)
        kthLargest(root, k);
 
    return 0;
}

                    

Java

// Java code to find k'th largest element in BST
 
// A binary tree node
class Node {
 
    int data;
    Node left, right;
 
    Node(int d)
    {
        data = d;
        left = right = null;
    }
}
 
class BinarySearchTree {
 
    // Root of BST
    Node root;
 
    // Constructor
    BinarySearchTree()
    {
        root = null;
    }
     
    // function to insert nodes
    public void insert(int data)
    {
        this.root = this.insertRec(this.root, data);
    }
     
    /* A utility function to insert a new node
    with given key in BST */
    Node insertRec(Node node, int data)
    {  
        /* If the tree is empty, return a new node */
        if (node == null) {
            this.root = new Node(data);
            return this.root;
        }
 
        if (data == node.data) {
            return node;
        }
         
        /* Otherwise, recur down the tree */
        if (data < node.data) {
            node.left = this.insertRec(node.left, data);
        } else {
            node.right = this.insertRec(node.right, data);
        }
        return node;
    }
 
    // class that stores the value of count
    public class count {
        int c = 0;
    }
 
    // utility function to find kth largest no in
    // a given tree
    void kthLargestUtil(Node node, int k, count C)
    {
        // Base cases, the second condition is important to
        // avoid unnecessary recursive calls
        if (node == null || C.c >= k)
            return;
         
        // Follow reverse inorder traversal so that the
        // largest element is visited first
        this.kthLargestUtil(node.right, k, C);
         
        // Increment count of visited nodes
        C.c++;
         
        // If c becomes k now, then this is the k'th largest
        if (C.c == k) {
            System.out.println(k + "th largest element is " +
                                                 node.data);
            return;
        }
         
        // Recur for left subtree
        this.kthLargestUtil(node.left, k, C);
    }
 
    // Method to find the kth largest no in given BST
    void kthLargest(int k)
    {
        count c = new count(); // object of class count
        this.kthLargestUtil(this.root, k, c);
    }
 
    // Driver function
    public static void main(String[] args)
    {
        BinarySearchTree tree = new BinarySearchTree();
         
        /* Let us create following BST
              50
           /     \
          30      70
         /  \    /  \
       20   40  60   80 */
        tree.insert(50);
        tree.insert(30);
        tree.insert(20);
        tree.insert(40);
        tree.insert(70);
        tree.insert(60);
        tree.insert(80);
 
        for (int i = 1; i <= 7; i++) {
            tree.kthLargest(i);
        }
    }
}
 
// This code is contributed by Kamal Rawal

                    

Python3

# Python3 program to find k'th largest
# element in BST
 
class Node:
 
    # Constructor to create a new node
    def __init__(self, data):
        self.key = data
        self.left = None
        self.right = None
         
# A function to find k'th largest
# element in a given tree.
def kthLargestUtil(root, k, c):
     
    # Base cases, the second condition
    # is important to avoid unnecessary
    # recursive calls
    if root == None or c[0] >= k:
        return
 
    # Follow reverse inorder traversal
    # so that the largest element is
    # visited first
    kthLargestUtil(root.right, k, c)
 
    # Increment count of visited nodes
    c[0] += 1
 
    # If c becomes k now, then this is
    # the k'th largest
    if c[0] == k:
        print("K'th largest element is",
                               root.key)
        return
 
    # Recur for left subtree
    kthLargestUtil(root.left, k, c)
 
# Function to find k'th largest element
def kthLargest(root, k):
     
    # Initialize count of nodes
    # visited as 0
    c = [0]
 
    # Note that c is passed by reference
    kthLargestUtil(root, k, c)
 
# A utility function to insert a new
# node with given key in BST */
def insert(node, key):
     
    # If the tree is empty,
    # return a new node
    if node == None:
        return Node(key)
 
    # Otherwise, recur down the tree
    if key < node.key:
        node.left = insert(node.left, key)
    elif key > node.key:
        node.right = insert(node.right, key)
 
    # return the (unchanged) node pointer
    return node
 
# Driver Code
if __name__ == '__main__':
     
    # Let us create following BST
    #         50
    #     /     \
    #     30     70
    # / \ / \
    # 20 40 60 80 */
    root = None
    root = insert(root, 50)
    insert(root, 30)
    insert(root, 20)
    insert(root, 40)
    insert(root, 70)
    insert(root, 60)
    insert(root, 80)
 
    for k in range(1,8):
        kthLargest(root, k)
         
# This code is contributed by PranchalK

                    

C#

using System;
 
// C# code to find k'th largest element in BST
 
// A binary tree node
public class Node
{
 
    public int data;
    public Node left, right;
 
    public Node(int d)
    {
        data = d;
        left = right = null;
    }
}
 
public class BinarySearchTree
{
 
    // Root of BST
    public Node root;
 
    // Constructor
    public BinarySearchTree()
    {
        root = null;
    }
 
    // function to insert nodes
    public virtual void insert(int data)
    {
        this.root = this.insertRec(this.root, data);
    }
 
    /* A utility function to insert a new node 
    with given key in BST */
    public virtual Node insertRec(Node node, int data)
    {
        /* If the tree is empty, return a new node */
        if (node == null)
        {
            this.root = new Node(data);
            return this.root;
        }
 
        if (data == node.data)
        {
            return node;
        }
 
        /* Otherwise, recur down the tree */
        if (data < node.data)
        {
            node.left = this.insertRec(node.left, data);
        }
        else
        {
            node.right = this.insertRec(node.right, data);
        }
        return node;
    }
 
    // class that stores the value of count
    public class count
    {
        private readonly BinarySearchTree outerInstance;
 
        public count(BinarySearchTree outerInstance)
        {
            this.outerInstance = outerInstance;
        }
 
        internal int c = 0;
    }
 
    // utility function to find kth largest no in 
    // a given tree
    public virtual void kthLargestUtil(Node node, int k, count C)
    {
        // Base cases, the second condition is important to
        // avoid unnecessary recursive calls
        if (node == null || C.c >= k)
        {
            return;
        }
 
        // Follow reverse inorder traversal so that the
        // largest element is visited first
        this.kthLargestUtil(node.right, k, C);
 
        // Increment count of visited nodes
        C.c++;
 
        // If c becomes k now, then this is the k'th largest 
        if (C.c == k)
        {
            Console.WriteLine(k + "th largest element is " + node.data);
            return;
        }
 
        // Recur for left subtree
        this.kthLargestUtil(node.left, k, C);
    }
 
    // Method to find the kth largest no in given BST
    public virtual void kthLargest(int k)
    {
        count c = new count(this); // object of class count
        this.kthLargestUtil(this.root, k, c);
    }
 
    // Driver function
    public static void Main(string[] args)
    {
        BinarySearchTree tree = new BinarySearchTree();
 
        /* Let us create following BST
              50
           /     \
          30      70
         /  \    /  \
       20   40  60   80 */
        tree.insert(50);
        tree.insert(30);
        tree.insert(20);
        tree.insert(40);
        tree.insert(70);
        tree.insert(60);
        tree.insert(80);
 
        for (int i = 1; i <= 7; i++)
        {
            tree.kthLargest(i);
        }
    }
}
 
  // This code is contributed by Shrikant13

                    

Javascript

<script>
// javascript code to find k'th largest element in BST
 
// A binary tree node
class Node {
 
   constructor(d)
    {
        this.data = d;
        this.left = this.right = null;
    }
}
 
 
 
    // Root of BST
    var root = null;
 
    // Constructor
    
     
    // function to insert nodes
    function insert(data)
    {
        this.root = this.insertRec(this.root, data);
    }
     
    /* A utility function to insert a new node
    with given key in BST */
    function insertRec( node , data)
    {  
        /* If the tree is empty, return a new node */
        if (node == null) {
            this.root = new Node(data);
            return this.root;
        }
 
        if (data == node.data) {
            return node;
        }
         
        /* Otherwise, recur down the tree */
        if (data < node.data) {
            node.left = this.insertRec(node.left, data);
        } else {
            node.right = this.insertRec(node.right, data);
        }
        return node;
    }
 
    // class that stores the value of count
     class count {
        constructor(){this.c = 0;}
   
    }
 
    // utility function to find kth largest no in
    // a given tree
    function kthLargestUtil( node , k,  C)
    {
        // Base cases, the second condition is important to
        // avoid unnecessary recursive calls
        if (node == null || C.c >= k)
            return;
         
        // Follow reverse inorder traversal so that the
        // largest element is visited first
        this.kthLargestUtil(node.right, k, C);
         
        // Increment count of visited nodes
        C.c++;
         
        // If c becomes k now, then this is the k'th largest
        if (C.c == k) {
            document.write(k + "th largest element is " +
                                                 node.data+"<br/>");
            return;
        }
         
        // Recur for left subtree
        this.kthLargestUtil(node.left, k, C);
    }
 
    // Method to find the kth largest no in given BST
    function kthLargest(k)
    {
         c = new count(); // object of class count
        this.kthLargestUtil(this.root, k, c);
    }
 
    // Driver function
     
  
         
        /* Let us create following BST
              50
           /     \
          30      70
         /  \    /  \
       20   40  60   80 */
        insert(50);
        insert(30);
        insert(20);
        insert(40);
        insert(70);
        insert(60);
        insert(80);
 
        for (i = 1; i <= 7; i++) {
            kthLargest(i);
        }
 
// This code contributed by gauravrajput1
</script>

                    

Output
K'th largest element is 80
K'th largest element is 70
K'th largest element is 60
K'th largest element is 50
K'th largest element is 40
K'th largest element is 30
K'th largest element is 20







Complexity Analysis:  

  1. Time Complexity: O(n). 
    In worst case the code can traverse each and every node of the tree if the k given is equal to n (total number of nodes in the tree). Therefore overall time complexity is O(n).
  2. Auxiliary Space: O(h). 
    Max recursion stack of height h at a given time.

This article is contributed by Chirag Sharma.

Approach 2:-  Iterative approach

The idea is to do reverse inorder traversal of BST. Keep a count of nodes visited.
The reverse inorder traversal traverses all nodes in decreasing order, i.e, visit the right node then centre then left and continue traversing the nodes recursively.
While doing the traversal, keep track of the count of nodes visited so far.
When the count becomes equal to k, stop the traversal and print the key. 

Time Complexity : – O(N)

C++

#include <iostream>
#include <stack>
 
using namespace std;
 
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
 
int kthLargest(TreeNode* root, int k) {
    stack<TreeNode*> st;
    TreeNode* curr = root;
    int count = 0;
    while (curr != NULL || !st.empty()) {
        while (curr != NULL) {
            st.push(curr);
            curr = curr->right;
        }
        curr = st.top();
        st.pop();
        count++;
        if (count == k) {
            return curr->val;
        }
        curr = curr->left;
    }
    return -1; // kth largest element does not exist
}
 
int main() {
    // create a BST
    TreeNode* root = new TreeNode(5);
    root->left = new TreeNode(3);
    root->right = new TreeNode(7);
    root->left->left = new TreeNode(2);
    root->left->right = new TreeNode(4);
    root->right->left = new TreeNode(6);
    root->right->right = new TreeNode(8);
 
    // find the kth largest element
    int k = 3;
    int kth_largest = kthLargest(root, k);
    if (kth_largest != -1) {
        cout << "The " << k << "th largest element is: " << kth_largest << endl;
    } else {
        cout << "The " << k << "th largest element does not exist" << endl;
    }
 
    return 0;
}

                    

Java

import java.util.Stack;
 
// Creating Tree
class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
 
    public TreeNode(int x) {
        val = x;
        left = null;
        right = null;
    }
}
 
// Function to find kth largest element
class KthLargestElementBST {
    public static int kthLargest(TreeNode root, int k) {
        Stack<TreeNode> stack = new Stack<>();
        TreeNode curr = root;
        int count = 0;
        while (curr != null || !stack.empty()) {
            while (curr != null) {
                stack.push(curr);
                curr = curr.right;
            }
            curr = stack.pop();
            count++;
            if (count == k) {
                return curr.val;
            }
            curr = curr.left;
        }
        return -1; // kth largest element does not exist
    }
 
  // Driver code
    public static void main(String[] args) {
        // create a BST
        TreeNode root = new TreeNode(5);
        root.left = new TreeNode(3);
        root.right = new TreeNode(7);
        root.left.left = new TreeNode(2);
        root.left.right = new TreeNode(4);
        root.right.left = new TreeNode(6);
        root.right.right = new TreeNode(8);
 
        // find the kth largest element
        int k = 3;
        int kth_largest = kthLargest(root, k);
        if (kth_largest != -1) {
            System.out.println("The " + k + "th largest element is: " + kth_largest);
        } else {
            System.out.println("The " + k + "th largest element does not exist");
        }
    }
}

                    

Python3

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
 
def kthLargest(root: TreeNode, k: int) -> int:
    st = []
    curr = root
    count = 0
    while curr or st:
        while curr:
            st.append(curr)
            curr = curr.right
        curr = st.pop()
        count += 1
        if count == k:
            return curr.val
        curr = curr.left
    return -1
 
# create a BST
root = TreeNode(5)
root.left = TreeNode(3)
root.right = TreeNode(7)
root.left.left = TreeNode(2)
root.left.right = TreeNode(4)
root.right.left = TreeNode(6)
root.right.right = TreeNode(8)
 
# find the kth largest element
k = 3
kth_largest = kthLargest(root, k)
if kth_largest != -1:
    print(f"The {k}th largest element is: {kth_largest}")
else:
    print(f"The {k}th largest element does not exist")

                    

C#

using System;
using System.Collections.Generic;
 
public class TreeNode {
    public int val;
    public TreeNode left;
    public TreeNode right;
 
    public TreeNode(int x)
    {
        val = x;
        left = null;
        right = null;
    }
}
 
public class KthLargestElementBST {
    // Function to find kth largest element
    public static int KthLargest(TreeNode root, int k)
    {
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode curr = root;
        int count = 0;
        while (curr != null || stack.Count > 0) {
            while (curr != null) {
                stack.Push(curr);
                curr = curr.right;
            }
            curr = stack.Pop();
            count++;
            if (count == k) {
                return curr.val;
            }
            curr = curr.left;
        }
        return -1; // kth largest element does not exist
    }
 
    // Driver code
    public static void Main()
    {
        // create a BST
        TreeNode root = new TreeNode(5);
        root.left = new TreeNode(3);
        root.right = new TreeNode(7);
        root.left.left = new TreeNode(2);
        root.left.right = new TreeNode(4);
        root.right.left = new TreeNode(6);
        root.right.right = new TreeNode(8);
 
        // find the kth largest element
        int k = 3;
        int kth_largest = KthLargest(root, k);
        if (kth_largest != -1) {
            Console.WriteLine("The " + k
                              + "th largest element is: "
                              + kth_largest);
        }
        else {
            Console.WriteLine(
                "The " + k
                + "th largest element does not exist");
        }
    }
}

                    

Javascript

class TreeNode {
  constructor(x) {
    this.val = x;
    this.left = null;
    this.right = null;
  }
}
 
function kthLargest(root, k) {
  const st = [];
  let curr = root;
  let count = 0;
  while (curr || st.length) {
    while (curr) {
      st.push(curr);
      curr = curr.right;
    }
    curr = st.pop();
    count++;
    if (count === k) {
      return curr.val;
    }
    curr = curr.left;
  }
  return -1;
}
 
// create a BST
const root = new TreeNode(5);
root.left = new TreeNode(3);
root.right = new TreeNode(7);
root.left.left = new TreeNode(2);
root.left.right = new TreeNode(4);
root.right.left = new TreeNode(6);
root.right.right = new TreeNode(8);
 
// find the kth largest element
const k = 3;
const kth_largest = kthLargest(root, k);
if (kth_largest !== -1) {
  console.log(`The ${k}th largest element is: ${kth_largest}`);
} else {
  console.log(`The ${k}th largest element does not exist`);
}

                    

Output
The 3th largest element is: 6







 Time complexity  –The time complexity of the kthLargest function is O(h + k), where h is the height of the binary search tree and k is the given value for finding the kth largest element. This is because in the worst case, the function will need to traverse the height of the tree and then traverse k elements to find the kth largest element.

Space complexity –The space complexity of the kthLargest function is O(h), where h is the height of the binary search tree. This is because the function uses a stack to store the nodes in the right subtree, which can have at most h nodes if the tree is skewed.

Using a Max Heap:

Follow the steps to implement the above approach:

  1. Create an empty max heap of integers using the C++ STL priority_queue container.
  2. Traverse the binary search tree in reverse in-order sequence using a stack.
  3. At each node, add the node value to the max heap.
  4. If the size of the max heap becomes greater than K, remove the maximum element from the max heap.
  5. After we finish traversing the binary search tree, the maximum element in the max heap will be the Kth largest element in the binary search tree.

Below is the implementation of the above approach:

C++

// C++ code to implement maxHeap approach
#include <bits/stdc++.h>
using namespace std;
 
// Definition of a binary tree node
struct Node {
    int data;
    Node *left, *right;
    Node(int val)
    {
        data = val;
        left = right = NULL;
    }
};
// Function to implement maxHeap approach
int kthLargest(Node* root, int K)
{
    priority_queue<int> maxHeap;
    // Traverse the binary search tree and add elements to
    // the max heap
    stack<Node*> s;
    Node* curr = root;
    while (curr != NULL || s.empty() == false) {
        while (curr != NULL) {
            s.push(curr);
            curr = curr->right;
        }
        curr = s.top();
        s.pop();
        maxHeap.push(curr->data);
        curr = curr->left;
    }
    // Remove K-1 elements from the max heap
    for (int i = 1; i < K; i++) {
        maxHeap.pop();
    }
    // The top element of the max heap is the Kth largest
    // element
    return maxHeap.top();
}
// Driver Code
int main()
{
    // Example binary search tree
    /*
              4
            /   \
           2     9
                / \
               7  10
    */
    Node* root = new Node(4);
    root->left = new Node(2);
    root->right = new Node(9);
    root->right->left = new Node(7);
    root->right->right = new Node(10);
 
    int K = 2;
    int kthLargestElement = kthLargest(root, K);
 
    cout << "The " << K
         << "th largest element in the binary search tree "
            "is: "
         << kthLargestElement << endl;
    return 0;
}
// This code is contributed by Veerendra_Singh_Rajpoot

                    

Java

import java.util.PriorityQueue;
import java.util.Stack;
 
// Definition of a binary tree node
class Node {
    int data;
    Node left, right;
 
    Node(int val) {
        data = val;
        left = right = null;
    }
}
 
public class KthLargestElement {
 
    // Function to implement maxHeap approach to find the Kth largest element
    public static int kthLargest(Node root, int K) {
        // Create a max heap to store the largest elements
        PriorityQueue<Integer> maxHeap = new PriorityQueue<>((a, b) -> b - a);
 
        // Traverse the binary search tree and add elements to the max heap
        Stack<Node> stack = new Stack<>();
        Node curr = root;
        while (curr != null || !stack.isEmpty()) {
            while (curr != null) {
                stack.push(curr);
                curr = curr.right;
            }
            curr = stack.pop();
            maxHeap.add(curr.data);
            curr = curr.left;
        }
 
        // Remove K-1 elements from the max heap
        for (int i = 1; i < K; i++) {
            maxHeap.poll();
        }
 
        // The top element of the max heap is the Kth largest element
        return maxHeap.peek();
    }
 
    // Driver Code
    public static void main(String[] args) {
        // Example binary search tree
        /*
                  4
                /   \
               2     9
                    / \
                   7  10
        */
        Node root = new Node(4);
        root.left = new Node(2);
        root.right = new Node(9);
        root.right.left = new Node(7);
        root.right.right = new Node(10);
 
        int K = 2;
        int kthLargestElement = kthLargest(root, K);
 
        System.out.println("The " + K + "th largest element in the binary search tree is: " + kthLargestElement);
    }
}

                    

Python3

import heapq
 
# Definition of a binary tree node
class Node:
    def __init__(self, val):
        self.data = val
        self.left = None
        self.right = None
 
# Function to implement maxHeap approach
def kthLargest(root, K):
    maxHeap = []
    # Traverse the binary search tree and add elements to
    # the max heap
    stack = []
    curr = root
    while curr or stack:
        while curr:
            stack.append(curr)
            curr = curr.right
        curr = stack.pop()
        heapq.heappush(maxHeap, -curr.data)
        curr = curr.left
    # Remove K-1 elements from the max heap
    for i in range(K - 1):
        heapq.heappop(maxHeap)
    # The top element of the max heap is the Kth largest
    # element
    return -maxHeap[0]
 
# Driver Code
if __name__ == '__main__':
    # Example binary search tree
    '''
              4
            /   \
           2     9
                / \
               7  10
    '''
    root = Node(4)
    root.left = Node(2)
    root.right = Node(9)
    root.right.left = Node(7)
    root.right.right = Node(10)
 
    K = 2
    kthLargestElement = kthLargest(root, K)
 
    print(f"The {K}th largest element in the binary search tree is: {kthLargestElement}")

                    

C#

// C# code to implement maxHeap approach
using System;
using System.Collections.Generic;
 
// Definition of a binary tree node
class Node
{
    public int Data;
    public Node Left, Right;
 
    public Node(int val)
    {
        Data = val;
        Left = Right = null;
    }
}
 
class GFG
{
    // Function to implement maxHeap approach
    static int KthLargest(Node root, int K)
    {
        // Create a max heap
        List<int> maxHeap = new List<int>();
 
        // Traverse the binary search tree and add elements to the max heap
        Stack<Node> stack = new Stack<Node>();
        Node curr = root;
 
        while (curr != null || stack.Count > 0)
        {
            while (curr != null)
            {
                stack.Push(curr);
                curr = curr.Right;
            }
            curr = stack.Pop();
            maxHeap.Add(curr.Data);
            curr = curr.Left;
        }
 
        // Sort the max heap in descending order
        maxHeap.Sort((x, y) => y.CompareTo(x));
 
        // The Kth largest element is at index K-1
        return maxHeap[K - 1];
    }
 
    // Driver Code
    static void Main(string[] args)
    {
        // Example binary search tree
        /*
                  4
                /   \
               2     9
                    / \
                   7  10
        */
        Node root = new Node(4);
        root.Left = new Node(2);
        root.Right = new Node(9);
        root.Right.Left = new Node(7);
        root.Right.Right = new Node(10);
 
        int K = 2;
        int kthLargestElement = KthLargest(root, K);
 
        Console.WriteLine("The " + K + "th largest element in the binary search tree is: " + kthLargestElement);
    }
}
 
// This code is contributed by Susobhan Akhuli

                    

Javascript

// Definition of a binary tree node
class Node {
    constructor(val) {
        this.data = val;
        this.left = null;
        this.right = null;
    }
}
 
// Function to implement maxHeap approach
function kthLargest(root, K) {
    const maxHeap = new MaxHeap();
 
    // Traverse the binary search tree and add elements to the max heap
    const stack = [];
    let curr = root;
    while (curr !== null || stack.length > 0) {
        while (curr !== null) {
            stack.push(curr);
            curr = curr.right;
        }
        curr = stack.pop();
        maxHeap.push(curr.data);
        curr = curr.left;
    }
 
    // Remove K-1 elements from the max heap
    for (let i = 1; i < K; i++) {
        maxHeap.pop();
    }
 
    // The top element of the max heap is the Kth largest element
    return maxHeap.top();
}
 
// MaxHeap implementation
class MaxHeap {
    constructor() {
        this.heap = [];
    }
 
    push(val) {
        this.heap.push(val);
        this.heapifyUp(this.heap.length - 1);
    }
 
    pop() {
        if (this.heap.length === 0) {
            return null;
        }
        if (this.heap.length === 1) {
            return this.heap.pop();
        }
 
        const root = this.heap[0];
        this.heap[0] = this.heap.pop();
        this.heapifyDown(0);
        return root;
    }
 
    top() {
        if (this.heap.length === 0) {
            return null;
        }
        return this.heap[0];
    }
 
    heapifyUp(index) {
        while (index > 0) {
            const parentIndex = Math.floor((index - 1) / 2);
            if (this.heap[index] <= this.heap[parentIndex]) {
                break;
            }
            [this.heap[index], this.heap[parentIndex]] = [this.heap[parentIndex], this.heap[index]];
            index = parentIndex;
        }
    }
 
    heapifyDown(index) {
        while (true) {
            const leftChildIndex = 2 * index + 1;
            const rightChildIndex = 2 * index + 2;
            let largestIndex = index;
 
            if (leftChildIndex < this.heap.length && this.heap[leftChildIndex] >
            this.heap[largestIndex]) {
                largestIndex = leftChildIndex;
            }
 
            if (rightChildIndex < this.heap.length && this.heap[rightChildIndex] >
             this.heap[largestIndex]) {
                largestIndex = rightChildIndex;
            }
 
            if (largestIndex === index) {
                break;
            }
 
            [this.heap[index], this.heap[largestIndex]] =
            [this.heap[largestIndex], this.heap[index]];
            index = largestIndex;
        }
    }
}
 
// Driver Code
// Example binary search tree
/*
          4
        /   \
       2     9
            / \
           7  10
*/
const root = new Node(4);
root.left = new Node(2);
root.right = new Node(9);
root.right.left = new Node(7);
root.right.right = new Node(10);
 
const K = 2;
const kthLargestElement = kthLargest(root, K);
 
console.log(`The ${K}th largest element in the binary search tree is: ${kthLargestElement}`);
// This code is contributed by Veerendra_Singh_Rajpoot

                    

Output
The 2th largest element in the binary search tree is: 9







Time Complexity: O(n logk) , The worst-case time complexity of building a max heap of size k is O(klogk).
The worst-case time complexity of traversing a binary search tree in reverse in-order sequence is O(n), where n is the number of nodes in the binary search tree.

Space Complexity: O(k), We use a max heap of size k to keep track of the k largest elements in the binary search tree.
Therefore, the overall space complexity of this approach is O(k).



Last Updated : 29 Oct, 2023
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