Recursive output all the legitimate sequences of a given n pair of braces

Similar topics are thought to be solved by recursive methods.

Solution: The legal sequence of parentheses, the number of open parentheses already inserted is greater than the number of closing parentheses.

(1) Inserting the opening parenthesis: the remaining brackets, and the left parenthesis;

(2) Insert closing parenthesis: The remaining brackets, the number of closing brackets is greater than the number of left parenthesis;

Code:

public class Parentheses {public
	static void Printpar (int l, int r, char[] str, int count) {
		if (l<0 | | r<l) return;
		if (l==0&&r==0) {
			System.out.println (str);
		} else{
			if (l>0) {
				str[count]= ' (';
				Printpar (L-1, R, str, count+1);
			if (r>l) {
				str[count]= ') ';
				Printpar (l, r-1, str, count+1);
	
	}} public static void Main (string[] args) {
		int count = 4;
		Char str[] = new char[count*2];
		Printpar (count, Count, str, 0);
	}

The results are as follows:

(((())))
((()()))
((())())
((()))()
(()(()))
(()()())
(()())()
(())(())
(())()()
()((()))
()(()())
()(())()
()()(())
()()()()