A binary tree is a tree which is characterized by one of the following properties:
- It can be empty (null).
- It contains a root node only.
- It contains a root node with a left subtree, a right subtree, or both. These subtrees are also binary trees.
In-order traversal is performed as
- Traverse the left subtree.
- Visit root.
- Traverse the right subtree.
For this in-order traversal, start from the left child of the root node and keep exploring the left subtree until you reach a leaf. When you reach a leaf, back up to its parent, check for a right child and visit it if there is one. If there is not a child, you've explored its left and right subtrees fully. If there is a right child, traverse its left subtree then its right in the same manner. Keep doing this until you have traversed the entire tree. You will only store the values of a node as you visit when one of the following is true:
- it is the first node visited, the first time visited
- it is a leaf, should only be visited once
- all of its subtrees have been explored, should only be visited once while this is true
- it is the root of the tree, the first time visited
Swapping: Swapping subtrees of a node means that if initially node has left subtree L and right subtree R, then after swapping, the left subtree will be R and the right subtree, L.
For example, in the following tree, we swap children of node 1.
                                Depth
    1               1            [1]
   / \             / \
  2   3     ->    3   2          [2]
   \   \           \   \
    4   5           5   4        [3]
In-order traversal of left tree is 2 4 1 3 5 and of right tree is 3 5 1 2 4.
Swap operation:
We define depth of a node as follows:
- The root node is at depth 1.
- If the depth of the parent node is d, then the depth of current node will bed+1.
Given a tree and an integer, k, in one operation, we need to swap the subtrees of all the nodes at each depth h, where h ∈ [k, 2k, 3k,...].  In other words, if h is a multiple of k, swap the left and right subtrees of that level.
You are given a tree of n nodes where nodes are indexed from [1..n] and it is rooted at 1. You have to perform t swap operations on it, and after each swap operation print the in-order traversal of the current state of the tree.
Function Description
Complete the swapNodes function in the editor below. It should return a two-dimensional array where each element is an array of integers representing the node indices of an in-order traversal after a swap operation.
swapNodes has the following parameter(s): 
- indexes: an array of integers representing index values of each , beginning with , the first element, as the root. 
- queries: an array of integers, each representing a  value.
Input Format 
The first line contains n, number of nodes in the tree.   
Each of the next n lines contains two integers, a b, where a is the index of left child, and b is the index of right child of ith node.   
Note: -1 is used to represent a null node.   
The next line contains an integer, t, the size of . 
Each of the next t lines contains an integer , each being a value .  
Output Format 
For each k, perform the swap operation and store the indices of your in-order traversal to your result array.  After all swap operations have been performed, return your result array for printing.  
Constraints
- Either or
- Either or
- The index of a non-null child will always be greater than that of its parent.
Sample Input 0
3
2 3
-1 -1
-1 -1
2
1
1
Sample Output 0
3 1 2
2 1 3
Explanation 0
As nodes 2 and 3 have no children, swapping will not have any effect on them. We only have to swap the child nodes of the root node.
    1   [s]       1    [s]       1   
   / \      ->   / \        ->  / \  
  2   3 [s]     3   2  [s]     2   3
Note:  [s] indicates that a swap operation is done at this depth.
Sample Input 1
5
2 3
-1 4
-1 5
-1 -1
-1 -1
1
2
Sample Output 1
4 2 1 5 3
Explanation 1
Swapping child nodes of node 2 and 3 we get
    1                  1  
   / \                / \ 
  2   3   [s]  ->    2   3
   \   \            /   / 
    4   5          4   5  
Sample Input 2
11
2 3
4 -1
5 -1
6 -1
7 8
-1 9
-1 -1
10 11
-1 -1
-1 -1
-1 -1
2
2
4
Sample Output 2
2 9 6 4 1 3 7 5 11 8 10
2 6 9 4 1 3 7 5 10 8 11
Explanation 2
Here we perform swap operations at the nodes whose depth is either 2 or 4 for and then at nodes whose depth is 4 for .
         1                     1                          1             
        / \                   / \                        / \            
       /   \                 /   \                      /   \           
      2     3    [s]        2     3                    2     3          
     /      /                \     \                    \     \         
    /      /                  \     \                    \     \        
   4      5          ->        4     5          ->        4     5       
  /      / \                  /     / \                  /     / \      
 /      /   \                /     /   \                /     /   \     
6      7     8   [s]        6     7     8   [s]        6     7     8
 \          / \            /           / \              \         / \   
  \        /   \          /           /   \              \       /   \  
   9      10   11        9           11   10              9     10   11