386. Lexicographical Numbers

class Solution {
    public List<Integer> lexicalOrder(int n) {
        List<Integer> result = new ArrayList<>();
        if (n <= 0) {
            return result;
        }

        for (int i=1; i<10; i++) {
            dfs(i, n, result);
        }

        return result;
    }

    public void dfs(int cur, int n, List<Integer> res) {
        if (cur > n) {
            return;
        }

        res.add(cur);
        for (int i=0; i<10; i++) {
            int newCur = cur*10 + i;

            if (newCur > n) {
                return;
            }

            dfs(newCur, n, res);
        }
    }
}