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K’th Largest element in BST using constant extra space

  • Difficulty Level : Hard
  • Last Updated : 21 May, 2021

Given a binary search tree, task is to find Kth largest element in the binary search tree. 
Example: 
 

Input :  k = 3
         Root of following BST
            10
          /    \
         4      20
        /      /   \
       2     15     40
Output : 15

 

The idea is to use Reverse Morris Traversal which is based on Threaded Binary Trees. Threaded binary trees use the NULL pointers to store the successor and predecessor information which helps us to utilize the wasted memory by those NULL pointers.
The special thing about Morris traversal is that we can do Inorder traversal without using stack or recursion which saves us memory consumed by stack or recursion call stack.
Reverse Morris traversal is just the reverse of Morris traversal which is majorly used to do Reverse Inorder traversal with constant O(1) extra memory consumed as it does not uses any Stack or Recursion.
To find Kth largest element in a Binary search tree, the simplest logic is to do reverse inorder traversal and while doing reverse inorder traversal simply keep a count of number of Nodes visited. When the count becomes equal to k, we stop the traversal and print the data. It uses the fact that reverse inorder traversal will give us a list sorted in descending order. 
 

Algorithm

 

1) Initialize Current as root.
2) Initialize a count variable to 0.
3) While current is not NULL :
   3.1) If current has no right child
   a) Increment count and check if count is equal to K.
      1) If count is equal to K, simply return current 
         Node as it is the Kth largest Node.
   b) Otherwise, Move to the left child of current.

   3.2) Else, here we have 2 cases:
   a) Find the inorder successor of current Node. 
      Inorder successor is the left most Node 
      in the right subtree or right child itself.
   b) If the left child of the inorder successor is NULL:
      1) Set current as the left child of its inorder 
         successor.
      2) Move current Node to its right.
   c) Else, if the threaded link between the current Node 
      and it's inorder successor already exists :
      1) Set left pointer of the inorder successor as NULL.
      2) Increment count and check if count is equal to K.
           a) If count is equal to K, simply return current
              Node as it is the Kth largest Node.

      3) Otherwise, Move current to it's left child.

 



C++




// CPP code for finding K-th largest Node using O(1)
// extra memory and reverse Morris traversal.
#include <bits/stdc++.h>
using namespace std;
 
struct Node {
    int data;
    struct Node *left, *right;
};
 
// helper function to create a new Node
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->data = data;
    temp->right = temp->left = NULL;
    return temp;
}
 
Node* KthLargestUsingMorrisTraversal(Node* root, int k)
{
    Node* curr = root;
    Node* Klargest = NULL;
 
    // count variable to keep count of visited Nodes
    int count = 0;
 
    while (curr != NULL) {
        // if right child is NULL
        if (curr->right == NULL) {
 
            // first increment count and check if count = k
            if (++count == k)
                Klargest = curr;
 
            // otherwise move to the left child
            curr = curr->left;
        }
 
        else {
 
            // find inorder successor of current Node
            Node* succ = curr->right;
 
            while (succ->left != NULL && succ->left != curr)
                succ = succ->left;
 
            if (succ->left == NULL) {
 
                // set left child of successor to the
                // current Node
                succ->left = curr;
 
                // move current to its right
                curr = curr->right;
            }
 
            // restoring the tree back to original binary
            //  search tree removing threaded links
            else {
 
                succ->left = NULL;
 
                if (++count == k)
                    Klargest = curr;
 
                // move current to its left child
                curr = curr->left;
            }
        }
    }
 
    return Klargest;
}
 
int main()
{
    // Your C++ Code
    /* Constructed binary tree is
          4
        /   \
       2     7
     /  \   /  \
    1    3  6    10 */
 
    Node* root = newNode(4);
    root->left = newNode(2);
    root->right = newNode(7);
    root->left->left = newNode(1);
    root->left->right = newNode(3);
    root->right->left = newNode(6);
    root->right->right = newNode(10);
 
    cout << "Finding K-th largest Node in BST : "
         << KthLargestUsingMorrisTraversal(root, 2)->data;
 
    return 0;
}

Java




// Java Program for finding K-th largest Node using O(1)
// extra memory and reverse Morris traversal.
class GfG
{
 
static class Node
{
    int data;
    Node left, right;
}
 
// helper function to create a new Node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.right = null;
    temp.left = null;
    return temp;
}
 
static Node KthLargestUsingMorrisTraversal(Node root, int k)
{
    Node curr = root;
    Node Klargest = null;
 
    // count variable to keep count of visited Nodes
    int count = 0;
 
    while (curr != null)
    {
        // if right child is NULL
        if (curr.right == null)
        {
 
            // first increment count and check if count = k
            if (++count == k)
                Klargest = curr;
 
            // otherwise move to the left child
            curr = curr.left;
        }
 
        else
        {
 
            // find inorder successor of current Node
            Node succ = curr.right;
 
            while (succ.left != null && succ.left != curr)
                succ = succ.left;
 
            if (succ.left == null)
            {
 
                // set left child of successor to the
                // current Node
                succ.left = curr;
 
                // move current to its right
                curr = curr.right;
            }
 
            // restoring the tree back to original binary
            // search tree removing threaded links
            else
            {
 
                succ.left = null;
 
                if (++count == k)
                    Klargest = curr;
 
                // move current to its left child
                curr = curr.left;
            }
        }
    }
    return Klargest;
}
 
// Driver code
public static void main(String[] args)
{
    // Your Java Code
    /* Constructed binary tree is
        4
        / \
    2 7
    / \ / \
    1 3 6 10 */
 
    Node root = newNode(4);
    root.left = newNode(2);
    root.right = newNode(7);
    root.left.left = newNode(1);
    root.left.right = newNode(3);
    root.right.left = newNode(6);
    root.right.right = newNode(10);
 
    System.out.println("Finding K-th largest Node in BST : " +
                    KthLargestUsingMorrisTraversal(root, 2).data);
}
}

Python3




# Python3 code for finding K-th largest
# Node using O(1) extra memory and
# reverse Morris traversal.
 
# helper function to create a new Node
class newNode:
    def __init__(self, data):
        self.data = data
        self.right = self.left = None
 
def KthLargestUsingMorrisTraversal(root, k):
    curr = root
    Klargest = None
 
    # count variable to keep count
    # of visited Nodes
    count = 0
 
    while (curr != None):
         
        # if right child is None
        if (curr.right == None):
 
            # first increment count and
            # check if count = k
            count += 1
            if (count == k):
                Klargest = curr
 
            # otherwise move to the left child
            curr = curr.left
 
        else:
 
            # find inorder successor of
            # current Node
            succ = curr.right
 
            while (succ.left != None and
                   succ.left != curr):
                succ = succ.left
 
            if (succ.left == None):
 
                # set left child of successor
                # to the current Node
                succ.left = curr
 
                # move current to its right
                curr = curr.right
 
            # restoring the tree back to 
            # original binary search tree
            # removing threaded links
            else:
 
                succ.left = None
                count += 1
                if (count == k):
                    Klargest = curr
 
                # move current to its left child
                curr = curr.left
 
    return Klargest
 
# Driver Code
if __name__ == '__main__':
 
    # Constructed binary tree is
    #     4
    #     / \
    # 2     7
    # / \ / \
    # 1 3 6 10
    root = newNode(4)
    root.left = newNode(2)
    root.right = newNode(7)
    root.left.left = newNode(1)
    root.left.right = newNode(3)
    root.right.left = newNode(6)
    root.right.right = newNode(10)
 
    print("Finding K-th largest Node in BST : ",
           KthLargestUsingMorrisTraversal(root, 2).data)
 
# This code is contributed by PranchalK

C#




// C# Program for finding K-th largest Node using O(1)
// extra memory and reverse Morris traversal.
using System;
using System.Collections.Generic;
 
class GfG
{
 
public class Node
{
    public int data;
    public Node left, right;
}
 
// helper function to create a new Node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.right = null;
    temp.left = null;
    return temp;
}
 
static Node KthLargestUsingMorrisTraversal(Node root, int k)
{
    Node curr = root;
    Node Klargest = null;
 
    // count variable to keep count of visited Nodes
    int count = 0;
 
    while (curr != null)
    {
        // if right child is NULL
        if (curr.right == null)
        {
 
            // first increment count and check if count = k
            if (++count == k)
                Klargest = curr;
 
            // otherwise move to the left child
            curr = curr.left;
        }
 
        else
        {
 
            // find inorder successor of current Node
            Node succ = curr.right;
 
            while (succ.left != null && succ.left != curr)
                succ = succ.left;
 
            if (succ.left == null)
            {
 
                // set left child of successor to the
                // current Node
                succ.left = curr;
 
                // move current to its right
                curr = curr.right;
            }
 
            // restoring the tree back to original binary
            // search tree removing threaded links
            else
            {
 
                succ.left = null;
 
                if (++count == k)
                    Klargest = curr;
 
                // move current to its left child
                curr = curr.left;
            }
        }
    }
    return Klargest;
}
 
// Driver code
public static void Main(String[] args)
{
    // Your C# Code
    /* Constructed binary tree is
        4
        / \
    2 7
    / \ / \
    1 3 6 10 */
 
    Node root = newNode(4);
    root.left = newNode(2);
    root.right = newNode(7);
    root.left.left = newNode(1);
    root.left.right = newNode(3);
    root.right.left = newNode(6);
    root.right.right = newNode(10);
 
    Console.Write("Finding K-th largest Node in BST : " +
                    KthLargestUsingMorrisTraversal(root, 2).data);
}
}
 
// This code has been contributed by 29AjayKumar

Javascript




<script>
 
// JavaScript Program for finding
// K-th largest Node using O(1)
// extra memory and reverse
// Morris traversal.
 
 class Node
{
     constructor(){
     this.data = 0;
     this.left = null;
     this.right = null;
     }
}
 
// helper function to create a new Node
function newNode(data)
{
    var temp = new Node();
    temp.data = data;
    temp.right = null;
    temp.left = null;
    return temp;
}
 
function KthLargestUsingMorrisTraversal(root , k)
{
    var curr = root;
    var Klargest = null;
 
    // count variable to keep count of visited Nodes
    var count = 0;
 
    while (curr != null)
    {
        // if right child is NULL
        if (curr.right == null)
        {
 
            // first increment count and
            // check if count = k
            if (++count == k)
                Klargest = curr;
 
            // otherwise move to the left child
            curr = curr.left;
        }
 
        else
        {
 
            // find inorder successor of
            // current Node
            var succ = curr.right;
 
            while (succ.left != null && succ.left != curr)
                succ = succ.left;
 
            if (succ.left == null)
            {
 
                // set left child of successor to the
                // current Node
                succ.left = curr;
 
                // move current to its right
                curr = curr.right;
            }
 
            // restoring the tree back to original binary
            // search tree removing threaded links
            else
            {
 
                succ.left = null;
 
                if (++count == k)
                    Klargest = curr;
 
                // move current to its left child
                curr = curr.left;
            }
        }
    }
    return Klargest;
}
 
// Driver code
  
    // Your JavaScript Code
    /* Constructed binary tree is
        4
        / \
       2 7
    / \ / \
    1 3 6 10 */
 
     root = newNode(4);
    root.left = newNode(2);
    root.right = newNode(7);
    root.left.left = newNode(1);
    root.left.right = newNode(3);
    root.right.left = newNode(6);
    root.right.right = newNode(10);
 
    document.write("Finding K-th largest Node in BST : " +
               KthLargestUsingMorrisTraversal(root, 2).data);
 
// This code contributed by aashish1995
 
</script>
Output: 
Finding K-th largest Node in BST : 7

 

Time Complexity : O(n) 
Auxiliary Space : O(1)
 

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