Palindrome Partitioning I

class Solution {
    public List<List<String>> partition(String s) {
        List<List<String>> result = new ArrayList<>();
        if (s == null) {
            return result;
        }
        if (s.length() == 0) {
            result.add(new ArrayList<String>());
            return result;
        }

        backtrack(result, new ArrayList<String>(), s, 0);

        return result;
    }

    public void backtrack(List<List<String>> list, ArrayList<String> tempList, String s, int start) {
        if (start == s.length()) {
            list.add(new ArrayList<String>(tempList));
            return;
        }
        for (int i = start; i < s.length(); i++) {
            if (isPalindrome(s, start, i)) {
                tempList.add(s.substring(start, i+1));
                backtrack(list, tempList, s, i + 1);
                tempList.remove(tempList.size() - 1);
            }
        }
    }

    public boolean isPalindrome (String s, int start, int end) {
        if (start == end) {
            return true;
        }
        while (start < end) {
            if (s.charAt(start++) != s.charAt(end--)) {
                return false;
            }
        }
        return true;
    }
}