K’th smallest element in BST using O(1) Extra Space
Given a Binary Search Tree (BST) and a positive integer k, find the k’th smallest element in the Binary Search Tree.
For example, in the following BST, if k = 3, then output should be 10, and if k = 5, then output should be 14.
We have discussed two methods in this post and one method in this post. All of the previous methods require extra space. How to find the k’th largest element without extra space?
The idea is to use Morris Traversal. In this traversal, we first create links to Inorder successor and print the data using these links, and finally revert the changes to restore original tree. See this for more details.
Below is the implementation of the idea.
C++
// C++ program to find k'th largest element in BST#include<bits/stdc++.h>using namespace std;// A BST nodestruct Node{ int key; Node *left, *right;};// A function to findint KSmallestUsingMorris(Node *root, int k){ // Count to iterate over elements till we // get the kth smallest number int count = 0; int ksmall = INT_MIN; // store the Kth smallest Node *curr = root; // to store the current node while (curr != NULL) { // Like Morris traversal if current does // not have left child rather than printing // as we did in inorder, we will just // increment the count as the number will // be in an increasing order if (curr->left == NULL) { count++; // if count is equal to K then we found the // kth smallest, so store it in ksmall if (count==k) ksmall = curr->key; // go to current's right child curr = curr->right; } else { // we create links to Inorder Successor and // count using these links Node *pre = curr->left; while (pre->right != NULL && pre->right != curr) pre = pre->right; // building links if (pre->right==NULL) { //link made to Inorder Successor pre->right = curr; curr = curr->left; } // While breaking the links in so made temporary // threaded tree we will check for the K smallest // condition else { // Revert the changes made in if part (break link // from the Inorder Successor) pre->right = NULL; count++; // If count is equal to K then we found // the kth smallest and so store it in ksmall if (count==k) ksmall = curr->key; curr = curr->right; } } } return ksmall; //return the found value}// A utility function to create a new BST nodeNode *newNode(int item){ Node *temp = new Node; temp->key = item; temp->left = temp->right = NULL; return temp;}/* 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 functionsint 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); for (int k=1; k<=7; k++) cout << KSmallestUsingMorris(root, k) << " "; return 0;} |
Java
// Java program to find k'th largest element in BST import java.util.*;class GfG {// A BST node static class Node { int key; Node left, right; }// A function to find static int KSmallestUsingMorris(Node root, int k) { // Count to iterate over elements till we // get the kth smallest number int count = 0; int ksmall = Integer.MIN_VALUE; // store the Kth smallest Node curr = root; // to store the current node while (curr != null) { // Like Morris traversal if current does // not have left child rather than printing // as we did in inorder, we will just // increment the count as the number will // be in an increasing order if (curr.left == null) { count++; // if count is equal to K then we found the // kth smallest, so store it in ksmall if (count==k) ksmall = curr.key; // go to current's right child curr = curr.right; } else { // we create links to Inorder Successor and // count using these links Node pre = curr.left; while (pre.right != null && pre.right != curr) pre = pre.right; // building links if (pre.right== null) { //link made to Inorder Successor pre.right = curr; curr = curr.left; } // While breaking the links in so made temporary // threaded tree we will check for the K smallest // condition else { // Revert the changes made in if part (break link // from the Inorder Successor) pre.right = null; count++; // If count is equal to K then we found // the kth smallest and so store it in ksmall if (count==k) ksmall = curr.key; curr = curr.right; } } } return ksmall; //return the found value } // A utility function to create a new BST node static Node newNode(int item) { Node temp = new Node(); temp.key = item; temp.left = null; temp.right = null; return temp; } /* A utility function to insert a new node with given key in BST */static 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 public static void main(String[] args) { /* 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); for (int k=1; k<=7; k++) System.out.print(KSmallestUsingMorris(root, k) + " "); }} |
Python3
# Python 3 program to find k'th # largest element in BST# A BST node 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 def KSmallestUsingMorris(root, k): # Count to iterate over elements # till we get the kth smallest number count = 0 ksmall = -9999999999 # store the Kth smallest curr = root # to store the current node while curr != None: # Like Morris traversal if current does # not have left child rather than # printing as we did in inorder, we # will just increment the count as the # number will be in an increasing order if curr.left == None: count += 1 # if count is equal to K then we # found the kth smallest, so store # it in ksmall if count == k: ksmall = curr.key # go to current's right child curr = curr.right else: # we create links to Inorder Successor # and count using these links pre = curr.left while (pre.right != None and pre.right != curr): pre = pre.right # building links if pre.right == None: # link made to Inorder Successor pre.right = curr curr = curr.left # While breaking the links in so made # temporary threaded tree we will check # for the K smallest condition else: # Revert the changes made in if part # (break link from the Inorder Successor) pre.right = None count += 1 # If count is equal to K then we # found the kth smallest and so # store it in ksmall if count == k: ksmall = curr.key curr = curr.right return ksmall # return the found value # 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 Codeif __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): print(KSmallestUsingMorris(root, k), end = " ") # This code is contributed by PranchalK |
C#
// C# program to find k'th largest element in BST using System;class GfG {// A BST node public class Node { public int key; public Node left, right; }// A function to find static int KSmallestUsingMorris(Node root, int k) { // Count to iterate over elements till we // get the kth smallest number int count = 0; int ksmall = int.MinValue; // store the Kth smallest Node curr = root; // to store the current node while (curr != null) { // Like Morris traversal if current does // not have left child rather than printing // as we did in inorder, we will just // increment the count as the number will // be in an increasing order if (curr.left == null) { count++; // if count is equal to K then we found the // kth smallest, so store it in ksmall if (count==k) ksmall = curr.key; // go to current's right child curr = curr.right; } else { // we create links to Inorder Successor and // count using these links Node pre = curr.left; while (pre.right != null && pre.right != curr) pre = pre.right; // building links if (pre.right == null) { // link made to Inorder Successor pre.right = curr; curr = curr.left; } // While breaking the links in so made temporary // threaded tree we will check for the K smallest // condition else { // Revert the changes made in if part (break link // from the Inorder Successor) pre.right = null; count++; // If count is equal to K then we found // the kth smallest and so store it in ksmall if (count == k) ksmall = curr.key; curr = curr.right; } } } return ksmall; //return the found value } // A utility function to create a new BST node static Node newNode(int item) { Node temp = new Node(); temp.key = item; temp.left = null; temp.right = null; return temp; } /* A utility function to insert a new node with given key in BST */static 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 public static void Main(String[] args) { /* 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); for (int k = 1; k <= 7; k++) Console.Write(KSmallestUsingMorris(root, k) + " "); }} // This code has been contributed by 29AjayKumar |
Javascript
<script>// javascript program to find k'th largest element in BST // A BST node class Node { constructor() { this.key = 0; this.left = null; this.right = null; } } // A function to find function KSmallestUsingMorris(root , k) { // Count to iterate over elements till we // get the kth smallest number var count = 0; var ksmall = Number.MIN_VALUE; // store the Kth smallest var curr = root; // to store the current node while (curr != null) { // Like Morris traversal if current does // not have left child rather than printing // as we did in inorder, we will just // increment the count as the number will // be in an increasing order if (curr.left == null) { count++; // if count is equal to K then we found the // kth smallest, so store it in ksmall if (count==k) ksmall = curr.key; // go to current's right child curr = curr.right; } else { // we create links to Inorder Successor and // count using these links var pre = curr.left; while (pre.right != null && pre.right != curr) pre = pre.right; // building links if (pre.right== null) { //link made to Inorder Successor pre.right = curr; curr = curr.left; } // While breaking the links in so made temporary // threaded tree we will check for the K smallest // condition else { // Revert the changes made in if part (break link // from the Inorder Successor) pre.right = null; count++; // If count is equal to K then we found // the kth smallest and so store it in ksmall if (count==k) ksmall = curr.key; curr = curr.right; } } } return ksmall; //return the found value } // A utility function to create a new BST node function newNode(item) { var temp = new Node(); temp.key = item; temp.left = null; temp.right = null; return temp; } /* A utility function to insert a new node with given key in BST */function insert(node , 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 /* Let us create following BST 50 / \ 30 70 / \ / \ 20 40 60 80 */ var root = null; root = insert(root, 50); insert(root, 30); insert(root, 20); insert(root, 40); insert(root, 70); insert(root, 60); insert(root, 80); for (k=1; k<=7; k++) document.write(KSmallestUsingMorris(root, k) + " "); // This code contributed by Rajput-Ji</script> |
Output
20 30 40 50 60 70 80
Time Complexity: O(n) where n is the size of BST
Auxiliary Space: O(1)
This article is contributed by Abhishek Somani. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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