



The only line of input contains the binary tree node elements in the level order form. The values of nodes are 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
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).
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
The output consists of a single line containing "Symmetric" if the given tree is symmetric, else "Asymmetric".
You are not required to print the expected output; it has already been taken care of, Just implement the function.
0 <= N <= 10^5
1 <= Data <= 10^5
Where 'N' denotes the number of nodes in the given binary tree and 'Data' denotes the node value.
Time limit: 1sec
We do two inorder traversals recursively of the tree simultaneously. For the first traversal, we traverse the left child first and then the right child, and in the second traversal, we traverse the right child first and then the left child. Let cur1 and cur2 be the current nodes of the first and second traversals respectively and initially both of them are passed as root. Then the algorithm follows: