Python海龟递归树

我想用python的turtle编写一个程序，创建一个具有级别的树。下面是一些I/O，您可以看到它应该做什么

我的程序对第一种情况有效，但对第二种情况打印太多。该计划的规定如下：

• 必须是递归的

• 只能使用以下功能：

  turtle.forward(100)     <-- turtle goes forward 100 steps
  turtle.right(90)        <-- turtle turns right 90 degrees
  turtle.penup()          <-- turtle lifts its pen up off of the paper
  turtle.forward(100)     <-- turtle goes forward 100 steps
  turtle.pendown()        <-- turtle puts its pen down on the paper
  turtle.pencolor("red")  <-- turtle uses red pen
  turtle.circle(100)      <-- turtle draws circle of radius 100 
  turtle.pencolor("blue") <-- turtle changes to blue pen (most other common colors work too!)
  turtle.forward(50)      <-- turtle moves forward 50 steps
  turtle.xcor()           <-- turtle returns its current x-coordinate
  turtle.ycor()           <-- turtle returns its current y-coordinate
  

  ---

  turtle.forward（100）我认为有两个问题
  
  首先，对于递归调用，第二个参数应该是n-1，而不是length/n。如果您正在绘制标高n，下一次调用将绘制标高n-1，而不是标高长度/n
  
  第二个问题是逃逸条件。在第一次更改中，当没有更多标高可供绘制时，或n==1时，绘制将完成
  
  这听起来像是家庭作业，所以我不会发布确切的代码。
  我不知道turtle是内置在python中的。你绝对让我开心。
  import turtle
  
  def tree(length,n):
      """ paints a branch of a tree with 2 smaller branches, like an Y"""
      if length < (length/n):
             return       # escape the function
      turtle.forward(length)        # paint the thik branch of the tree
      turtle.left(45)          # rotate left for smaller "fork" branch
      tree(length * 0.5,length/n)      # create a smaller branch with 1/2 the lenght of the parent branch
      turtle.right(90)         # rotoate right for smaller "fork" branch
      tree(length * 0.5,length/n)      # create second smaller branch
      turtle.left(45)          # rotate back to original heading
      turtle.backward(length)       # move back to original position
      return              # leave the function, continue with calling program