Question

我只是一个初学者，我试图通过制作一个递归函数来使用python的乌龟制作一个分形树。我认为我到目前为止已经有了这个功能，它使得我的分形的整个左侧，但它没有做出正确的一面。我该如何解决这个问题？任何和所有的建议将不胜感激！

def svTree( trunkLength, levels ):
    """ uses the turtle drawing functions to return a tree with a specified number of levels
    input: two integers, trunkLength and levels
    """

    newtrunkLength = trunkLength *.5


    forward(trunkLength)

    if levels ==1:
           penup()


    else:
        left(45)         
        forward(trunkLength * 0.5)
        backward(trunkLength * 0.5)
        right(90)
        forward(trunkLength * 0.5)
        backward(trunkLength * 0.5)
        left(90)
        return svTree(newtrunkLength,levels-1)

Answer 1

在递归左侧分支之前保存乌龟的位置，然后重置乌龟并绘制右侧分支：

def svTree(length, level):
   if level == 1:
      return

   #draw level's trunk
   down()
   forward(length)
   up()

   #save turtle position and heading
   pos = position()
   hdg = heading()

   #draw left branch
   left(45)
   svTree(length / 2, level - 1)

   #restore turtle position and heading
   setposition(pos)
   setheading(hdg)

   #draw right branch
   right(45)
   svTree(length / 2, level - 1)

如果您不想在每次方法调用时保存位置和标题，您还可以确保乌龟光标在每个方法调用开始时的结尾和标题处结束。

def svTree(length, level):
    if level == 1:
        return

    #draw level's trunk
    down()
    forward(length)
    up()

    #draw left branch
    left(45)
    svTree(length/2, level-1)

    #draw right branch
    right(90)
    svTree(length/2, level-1)

    #return cursor to state when function called
    left(45)
    backward(length)

在python中使用龟的分形树的右侧

1 个答案: