递归如何遍历二叉树并计算列表节点的数量？

Question

我无法通过以下 URL (https://repl.it/@flowerplowedup/SquigglyVerifiablePaint#main.c) 可视化这些以下打印和大小函数：

Answer 1

假设您输入了输入序列

4 8 2 7 9 1 3 -1

你最终得到一棵看起来像这样的树：

                           4
                          / \
                         /   \
                        2     8
                       / \   / \
                      1   3 7   9

print_ascending 的执行看起来像这样：

print_ascending( node(4) )
{
  node(4) != NULL
  print_ascending( node(4)->left == node(2) )
  {
    node(2) != NULL
    print_ascending( node(2)->left == node(1) )
    {
      node(1) != NULL
      print_ascending( node(1)->left == NULL )
      {
        NULL == NULL
        return
      }
      print( 1 )
      print_ascending( node(1)->right == NULL)
      {
        NULL == NULL
        return
      }
      return
    }
    print( 2 )
    print_ascending( node(2)->right == node(3))
    {
      node(3) != NULL
      print_ascending( node(3)->left == NULL )
      {
        NULL = NULL
        return
      }
      print( 3 )
      print_ascending( node(3)->right == NULL )
      {
        NULL = NULL
        return
      }
    }
    print( 4 )
    print_ascending( node(4)->right == node(8) )
    {
      node(8) != NULL
      print_ascending( node(8)->left == node(7) )
      {
        node(7) != NULL
        print_ascending( node(7)->left == NULL )
        {
          NULL == NULL
          return
        }
        print( 7 )
        print_ascending( node(7)->right == NULL )
        {
          NULL == NULL
          return
        }
        return
      }
      print( 8 )
      print_ascending( node(8)->right == node(9) )
      {
        node(9) != NULL 
        print_ascending( node(9)->left == NULL )
        {
          NULL == NULL
          return
        }
        print( 9 )
        print_ascending( node(9)->right == NULL )
        {
          NULL == NULL
          return
        }
        return
      }
      return
    }
    return
  }
  return
}
return

希望这有助于可视化正在发生的事情，以及 size 函数中正在发生的事情。递归是您需要一段时间才能理解的概念之一。