Sort the given LinkedList

problem

note: the sort() method can be used to merge to sorted linked list

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode sortList(ListNode head) {
        //lets solve it via merge sort
        return sort(head);
    }
    public ListNode getMiddle(ListNode node){
        if(node ==null) return null;
        ListNode slow = node;
        ListNode fast = node.next;
        while(fast!=null && fast.next!=null){ // important 
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
    public ListNode sort(ListNode node){
        if(node==null || node.next ==null) return node;// important base case

        ListNode middle = getMiddle(node);
        ListNode temp = middle.next;//right
        middle.next = null;//left
        ListNode left = sort(node);
        ListNode right = sort(temp);
        return merge(left,right);
    }
    public ListNode merge(ListNode a, ListNode b){
        ListNode head = new ListNode();
        ListNode temp = head;

        while(a!=null && b!=null){

            if(a.val<=b.val){
                head.next = new ListNode(a.val);
                a = a.next;
            }
            else{
                head.next = new ListNode(b.val); 
                b = b.next;  
            }
            head = head.next;
        }
        if(a!=null) head.next = a;
        if(b!=null) head.next = b;
        return temp.next;
    }
}