pth common ancestor in BST
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Given a Binary Search Tree and two node values x and y present in the BST. Your task is to find the pth (p >= 1) common ancestor of these two nodes x and y.

The 1st common ancestor is the lowest common ancestor. The lowest common ancestor between two nodes n1 and n2 is defined as the lowest node in T that has both n1 and n2 as descendants (where we allow a node to be a descendant of itself).

All the elements of BST are non-negative and there is no duplicate entry in BST. Also, x and y are unequal.

You are required to complete the function pthCommonAncestor().

Return -1 if the pth common ancestor doesn’t exist for a given BST.

Note: The Input/Ouput format and Example given are used for system's internal purpose, and should be used by a user for Expected Output only. As it is a function problem, hence a user should not read any input from stdin/console. The task is to complete the function specified, and not to write the full code.
 

Input:

The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consists of three lines. The first line of each test case consists of two integers N and p, denoting the number of elements in the BST and the pth common ancestor to be found, respectively. The second line of each test case consists of N space-separated integers denoting the elements in the BST. The third line of each test case consists of two integers x and y as described in the problem statement.


Output:

You are required to complete the function pthCommonAncestor() which takes the root of the tree and two integers x and y(as described in the problem statement) as the arguments. The function returns the value of pth common ancestor to be found.

 

Constraints:

 

1 <= T <= 1000

2 <= N <= 1000

1 <= p <= 100


Example:

Input:

2

6 2

50 30 70 60 55 65

55 65

5 2

6 3 8 1 4

1 4

Output:

70

6

Explanation:

The BST for the sample test case 1 is-

             50                // 50 is the 3rd common ancestor of 55 and 65

        /            \

   30                70     // 70 is the 2nd common ancestor of 55 and 65

                   /

              60              // 60 is the 1st common ancestor of 55 and 65

           /      \   

       55        65

Hence, the value of 2nd common ancestor of 55 and 65 is 70.

The BST for the sample test case 2 is-

                        6                // 6 is the 2nd common ancestor of 1 and 4

                  /            \

            3                   8       // 3 is the 1st common ancestor of 1 and 4

        /         \

   1                4

Hence, the value of 2nd common ancestor of 1 and 4 is 6.

 

 

** For More Input/Output Examples Use 'Expected Output' option **

Contributor: Sajal Agrawal
Author: sajalagrawal


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