Given a Binary Search Tree (BST) and a positive integer k, find the k’th largest element in the Binary Search Tree.
 /   / \   / \  
Method 1) Brute Force in O(n)
In order of BST gives 5,8,9,10,12,20,22
and print member A[n-k+1] will be out put.
Now Lets discuss tricky way :
As we know that larger elements are in Right Side of a node and smaller in left side for BST.
Can we use this trick ???
Think for above example , R=no if Recursion,
r = 1 we traverse 22 and found that its largest
r = 2 we traverse 20, which is second largest.
r=3 and we went to 12 via 8-10-12 ( first traverse right path always)
so 3rd largest is 12.
So with Post Order Traversal We can find Kth largest in O(k + h) time
So Method 2 ) Post Order Traversal
Algo . C=Count of node visited, K= Kth number
Function KthLargest(Node *root, int k, int &c)
- If root=null or c>= k
- KthLargest ( root->right , k, c)
- if ( c == k) cout << root -> n;
- KthLargest ( root->left , k, c)
For queries, suggestion and more questions / solutions please write us at email@example.com