Given a Binary Tree (BT), convert it to a Doubly Linked List(DLL) In-Place. The left and right pointers in nodes are to be used as previous and next pointers respectively in converted DLL. The order of nodes in DLL must be same as Inorder of the given Binary Tree. The first node of Inorder traversal (left most node in BT) must be head node of the DLL.

**Input Format:**

The input contains **T**, denoting number of testcases. For each testcase there will be two lines. The first line contains number of edges. The second line contains two nodes and a character separated by space. The first node denotes data value, second node denotes where it will be assigned to the previous node which will depend on character '**L**' or '**R**' i.e. the 2nd node will be assigned as left child to the 1st node if character is 'L' and so on. The first node of second line is **root node**. The struct or class **Node** has a data part which stores the data, pointer to left child and pointer to right child. There are multiple test cases. For each test case, the function will be called individually.

**Output Format:**

For each testcase, in a new line, print front to end and back to end traversals of DLL.

**Your Task:**

You don't have to take input. Complete the function **bToDLL() **that takes **node and head **as parameter. The driver code prints the DLL both ways.

**Constraints:**

1 <= T <= 100

1 <= Number of nodes <= 100

1 <= Data of a node <= 1000

**Example:
Input:**

2

2

1 2 R 1 3 L

4

10 20 L 10 30 R 20 40 L 20 60 R

**Output:**

3 1 2

2 1 3

40 20 60 10 30

30 10 60 20 40

**Explanation:**

**Testcase1:**** **The tree is

1

/ \

3 2

So, DLL would be 3<=>1<=>2

**Testcase2:** The tree is

10

/ \

20 30

/ \

40 60

So, DLL would be 40<=>20<=>60<=>10<=>30.

Author: kartik

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