# Question

Formatted question description: https://leetcode.ca/all/1265.html

```
1265. Print Immutable Linked List in Reverse
You are given an immutable linked list, print out all values of each node in reverse with the help of the following interface:
ImmutableListNode: An interface of immutable linked list, you are given the head of the list.
You need to use the following functions to access the linked list (you can't access the ImmutableListNode directly):
ImmutableListNode.printValue(): Print value of the current node.
ImmutableListNode.getNext(): Return the next node.
The input is only given to initialize the linked list internally. You must solve this problem without modifying the linked list.
In other words, you must operate the linked list using only the mentioned APIs.
Example 1:
Input: head = [1,2,3,4]
Output: [4,3,2,1]
Example 2:
Input: head = [0,-4,-1,3,-5]
Output: [-5,3,-1,-4,0]
Example 3:
Input: head = [-2,0,6,4,4,-6]
Output: [-6,4,4,6,0,-2]
Constraints:
The length of the linked list is between [1, 1000].
The value of each node in the linked list is between [-1000, 1000].
Follow up:
Could you solve this problem in:
Constant space complexity?
Linear time complexity and less than linear space complexity?
```

# Algorithm

Similar to 206 Reverse Linked List

DFS, recursion before print current node.

- space is O(N), since need to store previous recursion

### Follow up

ould you solve this problem in:

Constant space complexity? Linear time complexity and less than linear space complexity?

# Code

Java

```
class Solution(object):
def generateSentences(self, synonyms, text):
"""
:type synonyms: List[List[str]]
:type text: str
:rtype: List[str]
"""
text = '#' + text.replace(' ', '#') + '#'
queue = [text]
for (w1,w2) in synonyms:
for text in queue :
newtext = text.replace('#' + w1+'#','#' + w2+'#',1)
if newtext != text and newtext not in queue:
queue.append(newtext)
newtext = text.replace('#' + w2 + '#', '#' + w1 + '#',1)
if newtext != text and newtext not in queue:
queue.append(newtext)
res = []
for i in sorted(queue):
newtext = i.replace('#', ' ')
res.append(newtext[1:-1])
return res
```