Explain Morris inorder tree traversal without using stacks or recursion

Question

Can someone please help me understand the following Morris inorder tree traversal algorithm without using stacks or recursion ? I was trying to understand how it works, but

Answer 1

public static void morrisInOrder(Node root) {
        Node cur = root;
        Node pre;
        while (cur!=null){
            if (cur.left==null){
                System.out.println(cur.value);      
                cur = cur.right; // move to next right node
            }
            else {  // has a left subtree
                pre = cur.left;
                while (pre.right!=null){  // find rightmost
                    pre = pre.right;
                }
                pre.right = cur;  // put cur after the pre node
                Node temp = cur;  // store cur node
                cur = cur.left;  // move cur to the top of the new tree
                temp.left = null;   // original cur left be null, avoid infinite loops
            }        
        }
    }

I think this code would be better to understand, just use a null to avoid infinite loops, don't have to use magic else. It can be easily modified to preorder.