Kth Largest Element in BST

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Problem statement

Given the root node of a Binary Search Tree (BST), you have to return the Kth largest element in the BST.

For Example: 
If K is 4 and the tree is depicted by the following image then,

Example1

The 4th largest element in the given BST is 1. So the output will be 1.
Follow-up : 
 Try to do it in O(1) space without using recursion.
Detailed explanation ( Input/output format, Notes, Images )
Input Format : 
The first line contains an Integer 'T' which denotes the number of test cases or queries to be run. Then the test cases follow.

The first line of each test case contains the elements of the tree in the level order form separated by a single space.
If any node does not have a left or right child, take -1 in its place. Refer to the example below.

The second line of each test case contains a single integer ‘K’, denoting ‘K’ as explained in the problem statement.
Example : 
Elements are in the level order form. The input consists of values of nodes separated by a single space in a single line. In case a node is null, we take -1 in its place.

For example, the input for the tree depicted in the below image would be :


1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1

Explanation :
Level 1 :
The root node of the tree is 1

Level 2 :
Left child of 1 = 2
Right child of 1 = 3

Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6

Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)

Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)

The first not-null node (of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.

The input ends when all nodes at the last level are null (-1).

Note: The above format was just to provide clarity on how the input is formed for a given tree. 

The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:

1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
Output Format : 
For each test case, print a single integer value representing the Kth largest value of the given BST.

Output for each test case will be printed in a separate line.
Note: 
You are not required to print anything, it has already been taken care of. Just implement the function.
Constraints : 
1 <= T <= 100
1 <= N <= 5 * 10^3
1 <= K <= N
0 <= X <= 10^9

Where ‘X’ is the value at the node and ‘N’ is the no. of nodes in given BST.

Time Limit: 1 sec.
Sample Input 1 :
2
2 1 4 -1 -1 3 -1 -1 -1
4
2 1 7 -1 -1 3 -1 -1 4 -1 -1
3
Sample Output 1 :
1
3
Explanation For Sample Input 1 :
For First Case - Same as explained in above example.

For the second case - 
K is 3 and the tree is depicted by the following image then,

Example2

The 3rd largest element in the given BST is 3. So the output will be 3.
Sample Input 2 :
2
4 3 -1 2 -1 1 -1 -1 -1
1
3 1 8 -1 2 -1 -1 -1 -1
3
Sample Output 2 :
4
2
Hint

Can we get elements in sorted order by traversing efficiently?

Approaches (2)
Inorder Traversal

The main idea is to use inorder traversal and get elements of the BST in sorted order. Then return the (n-k)th element. as it will be the Kth largest element. The implementation details are as per the following algorithm.

 

Algorithm:

 

Inorder Function

Function arguments - Node * root, depicting the pointer to the node to which we are traversing, integer array order to store the elements in sorted order.

Saves the elements of given BST in sorted order in the order array.

  • If root == NULL then, return
  • inorder(root->left, order)
  • Append root->data to order
  • inorder(root->right, order)

 

Given Function

  • Declare integer array order
  • inorder(root, order)
  • Declare and initialize integer n = size of order
  • Return order[n-k]
Time Complexity

O(N) where 'N' is the number of nodes in the given BST.

 

We are doing a traversal over complete BST which will take O(N) time. So total time will be O(N).

Space Complexity

O(N) where 'N' is the number of nodes in the given BST.

 

We only need extra space for the order and recursion stack. These are at most ‘N’ in size and take O(N) storage. So total space required will be O(N) + O(N) = O(N).

Code Solution
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Kth Largest Element in BST
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