在不使用堆栈或递归的情况下解释Morris有序树遍历

有人可以在不使用堆栈或递归的情况下帮助我了解以下Morris有序树遍历算法吗？我试图理解它是如何工作的，但是它只是在逃避我。

 1. Initialize current as root
 2. While current is not NULL
  If current does not have left child     
   a. Print current’s data
   b. Go to the right, i.e., current = current->right
  Else
   a. In current's left subtree, make current the right child of the rightmost node
   b. Go to this left child, i.e., current = current->left

我知道树的修改方式current node是right child，将max nodein做成in，right subtree并使用此属性进行有序遍历。但是除此之外，我迷路了。

编辑：找到了此随附的c ++代码。我很难理解修改后的树是如何还原的。魔术在于else子句，一旦修改了正确的叶子，该子句就会被击中。有关详细信息，请参见代码：

/* Function to traverse binary tree without recursion and
   without stack */
void MorrisTraversal(struct tNode *root)
{
  struct tNode *current,*pre;

  if(root == NULL)
     return; 

  current = root;
  while(current != NULL)
  {
    if(current->left == NULL)
    {
      printf(" %d ", current->data);
      current = current->right;
    }
    else
    {
      /* Find the inorder predecessor of current */
      pre = current->left;
      while(pre->right != NULL && pre->right != current)
        pre = pre->right;

      /* Make current as right child of its inorder predecessor */
      if(pre->right == NULL)
      {
        pre->right = current;
        current = current->left;
      }

     // MAGIC OF RESTORING the Tree happens here: 
      /* Revert the changes made in if part to restore the original
        tree i.e., fix the right child of predecssor */
      else
      {
        pre->right = NULL;
        printf(" %d ",current->data);
        current = current->right;
      } /* End of if condition pre->right == NULL */
    } /* End of if condition current->left == NULL*/
  } /* End of while */
}