以前写过的回溯算法基本都是只需要一次回溯操作的，这道题思路不太一样，需要把左括号和右括号都分别回溯一次，剪枝条件是判断左括号数量大于n或右括号数量大于n或右括号数量大于左括号数量
代码如下：
class Solution {
private List<String> list = new ArrayList<>();
private StringBuilder path = new StringBuilder();
private void backtracking(int leftCount, int rightCount, int n) {
if (leftCount > n || rightCount > n ||rightCount > leftCount) return;
if (path.length() == 2 * n) {
list.add(new String(path));
return;
}
path.append("(");
backtracking(leftCount + 1, rightCount, n);
path.deleteCharAt(path.length() - 1);
path.append(")");
backtracking(leftCount, rightCount + 1, n);
path.deleteCharAt(path.length() - 1);
}
public List<String> generateParenthesis(int n) {
backtracking(0, 0, n);
return list;
}
}