```
public List<Integer> lexicalOrder(int n) {
List<Integer> list = new ArrayList<>(n);
int curr = 1;
for (int i = 1; i <= n; i++) {
list.add(curr);
if (curr * 10 <= n) {
curr *= 10;
} else if (curr % 10 != 9 && curr + 1 <= n) {
curr++;
} else {
while ((curr / 10) % 10 == 9) {
curr /= 10;
}
curr = curr / 10 + 1;
}
}
return list;
}
```

The basic idea is to find the next number to add.

Take 45 for example: if the current number is 45, the next one will be 450 (450 == 45 * 10)(if 450 <= n), or 46 (46 == 45 + 1) (if 46 <= n) or 5 (5 == 45 / 10 + 1)(5 is less than 45 so it is for sure less than n).

We should also consider n = 600, and the current number = 499, the next number is 5 because there are all "9"s after "4" in "499" so we should divide 499 by 10 until the last digit is not "9".

It is like a tree, and we are easy to get a sibling, a left most child and the parent of any node.