Leetcode Medium: Letter Combinations of a Phone Number

A few days back I coded a Leetcode medium problem: Letter Combinations of a Phone Number (problem #17) as follows:

import java.util.Hashtable;

class Solution {
    public List<String> letterCombinations(String digits) {
        ArrayDeque<String> res = new ArrayDeque<>();
        res.add("");
        
        Hashtable<Integer, String> ht = new Hashtable<>();
        ht.put(2, "abc");
        ht.put(3, "def");
        ht.put(4, "ghi");
        ht.put(5, "jkl");
        ht.put(6, "mno");
        ht.put(7, "pqrs");
        ht.put(8, "tuv");
        ht.put(9, "wxyz");
        
        letterCombinationsRecur(digits, 0, ht, res);
        return new ArrayList<String>(res);
    }
    
    private void letterCombinationsRecur(final String digits, int digPtr, final Hashtable<Integer, String> ht, final ArrayDeque<String> res) {
        
        if (digPtr==digits.length()) {
            if (res.peekFirst().equals(""))
                res.removeFirst();
            return;
        }
        
        int inputResCnt = res.size();
        ArrayList<String> resAL = new ArrayList<>(res);
        
        
        String newLastChars = ht.get(digits.charAt(digPtr) - '0');
        int nlcLen = newLastChars.length();
        for (int nlcIdx=0; nlcIdx<nlcLen; nlcIdx++) {
            
            for (int i=0; i<inputResCnt; i++) {
                res.add(resAL.get(i) + newLastChars.charAt(nlcIdx));
            }
            
        }
        
        for (int i=0; i<inputResCnt; i++) {
            res.removeFirst();
        }
        letterCombinationsRecur(digits, ++digPtr, ht, res);
        
    }
}

Today, I revisited that problem and thought I could use a simplified technique to solve the problem by modifying the above solution to rely less of Level Order BFS-like dequeue population and rely more over recursion. Here's the result that resultantly significantly simpler:

class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> result = new ArrayList<>();
        if (digits == null || digits.length() == 0)
            return result;
        
        String[] mapping = {"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        
        letterCombinationsRecur (digits, 0, "", mapping, result);
        return result;
    }
    
    void letterCombinationsRecur (String digits, int digitPtr, String currentLetterCombination, String[] mapping, List<String> result) {
        if (digitPtr == digits.length()) {
            result.add(currentLetterCombination);
            return;
        }
        
        String currDigLetters = mapping[digits.charAt(digitPtr)-'0'];
        for (int i=0; i<currDigLetters.length(); i++) {
            letterCombinationsRecur(digits, digitPtr+1, currentLetterCombination+currDigLetters.charAt(i), mapping, result);
        }
    }
    
}

Let me know your thoughts in the comment section below! Meanwhile, happy coding!!