How to check for unbalanced set of parentheses inside a node?

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How do I check for any unbalanced set of parentheses inside a node using LINQ-to-XML, for example

XML

<?xml version="1.0"?>
    <root>
    	<child1>Anything goes <a>(here</a>) and <bb>(there)</bb>!</child1>
    	<child2>Whatever <a>goes (around)</a> <bb>comes around()</bb> and (<a>that is)</a> a fact.</child1>
    </root>

In the above sample xml, the node <a>(here</a> contains only a opening parentheses inside the node and the closing parentheses is outside the node, in that case I want to re-position that opening parentheses to just before the <a> node as (<a>here</a>) and the same logic for any node like (<a>that is)</a> should be replaced as (<a>that is</a>) and <a>goes (around)</a> should remain unchanged as it contains a complete set of opening and closing parentheses, however <bb>(there)</bb> should also be changed to (<bb>there</bb>) as both the opening and closing parentheses are positioned immediately after and before the nodes.
Can anybody help!!

What I have tried:

I'm clueless how to do this to be honest

XDocument xml = XDocument.Load(@"D:\test.xml");
var targetnodes=from x in xml.Descendants("a")
                where ????

Also, here are some other types for better understanding
<bb>(statistical moments of the 1(st) - 4(th) (order))</bb>
(<bb>statistical moments of the 1(st) - 4(th) (order))</bb>
<bb>(statistical moments of the 1(st) - 4(th) (order)</bb>)
<bb>((statistical moments) of the 1(st) - 4(th) (order)</bb>)
should be
(<bb>statistical moments of the 1(st) - 4(th) (order)</bb>)
(<bb>statistical moments of the 1(st) - 4(th) (order)</bb>)
(<bb>statistical moments of the 1(st) - 4(th) (order)</bb>)
(<bb>(statistical moments) of the 1(st) - 4(th) (order)</bb>)

Posted 12-Mar-18 4:32am

Member 12692000

Updated 12-Mar-18 5:24am

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W∴ Balboos, *GHB* · Answer 1 · 2018-03-12T05:25:00

One strategy I used some time ago for similar types of things:

You need to push/pop things from a stack - actually, possibly more than one stack.

For elements: you push the name of the element on the stack. When a closing element comes up, compare it to the top of the stack and it must match (and if it does, pop it).
This is necessary so that paring occurs within an element.

Now, for the parenthesis: they, like braces, etc., must occur in pairs between elements (the push/pop above). If you use the same stack, whenever you come across a paren, compare it to the top element. If you've just gotten a closing paren then the top element should be an opening paren. If not, push the new paren (they're nesting). Eventually, you'll reach the closing element and there should be no paren's left. Same test as in paragraph, above.