Build Binary Expression Tree From Infix Expression

You are given a string ‘S’, an infix expression that has operands, operators and parentheses ‘(‘ and ‘)’.

You need to return the binary expression tree, whose inorder traversal is the same as ‘S’.

Note:

Infix expression: The expression of the form ‘a operator b’. When an operator is in-between every pair of operands.
The expression tree is a binary tree in which each internal node corresponds to the operator and each leaf node corresponds to the operand so for example expression tree for 5 * ( 6 - 3 ) / 2 - 8 would be:

Input Format:

The first line of input contains an integer 'T' representing the number of test cases.

The first line of each test case contains a string S representing infix expression.

Output Format :

For each test case, return the binary expression tree, whose inorder traversal is the same as 'S'. 

Return -1 instead of NULL.

The output for each test case is printed in a separate line.

Note:

You do not need to print anything, it has already been taken care of. Just implement the given function.

Constraints :

1 <= T <= 5
1 <= N <= 5000
Operands are only numbers between 0 and 9 (included).
It is guaranteed that ‘S’ is a valid infix expression.

Time limit: 1 sec