K’th Largest element in BST using constant extra space
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 <iostream> 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; } |
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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); } } |
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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 |
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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 |
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Output:
Finding K-th largest Node in BST : 7
Time Complexity : O(n)
Auxiliary Space : O(1)
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