在 Python 中绘制分形树，不知道如何进行

Question

到目前为止，我在 python 中有这个

import turtle
import math
t = turtle.Turtle()
t.shape("turtle")
t.lt(90)

lv = 11
l  = 100
s  = 17

t.penup()
t.bk(l)
t.pendown()
t.fd(l)

def draw_tree(l, level):
    l = 3.0/4.0*l
    t.lt(s)
    t.fd(l)
    level +=1
    if level<lv:
        draw_tree(l, level)

    t.bk(l)
    t.rt(2*s)
    t.fd(l)
    if level<=lv:
        draw_tree(l, level)
    t.bk(l)
    t.lt(s)
    level -=1

t.speed(100)        
draw_tree(l, 2)

但我有点被困在如何进步上，因为我需要伸手去建造这棵树。这就是我想要制作的：

谁能告诉我我做错了什么？

Answer 1

我真的很喜欢@cdlane 的回答，所以我玩了一段时间的代码。树现在看起来好多了，代码也更具可读性，所以我认为值得分享。

编码：

import turtle

WIDTH = 15
BRANCH_LENGTH = 120
ROTATION_LENGTH = 27


class Tree_Fractal(turtle.Turtle):
    def __init__(self, level):
        super(Tree_Fractal, self).__init__()
        self.level = level
        self.hideturtle()
        self.speed('fastest')
        self.left(90)
        self.width(WIDTH)
        self.penup()
        self.back(BRANCH_LENGTH * 1.5)
        self.pendown()
        self.forward(BRANCH_LENGTH)
        self.draw_tree(BRANCH_LENGTH, level)

    def draw_tree(self, branch_length, level):
        width = self.width()
        self.width(width * 3. / 4.)
        branch_length *= 3. / 4.
        self.left(ROTATION_LENGTH)
        self.forward(branch_length)

        if level > 0:
            self.draw_tree(branch_length, level - 1)
        self.back(branch_length)
        self.right(2 * ROTATION_LENGTH)
        self.forward(branch_length)

        if level > 0:
            self.draw_tree(branch_length, level - 1)
        self.back(branch_length)
        self.left(ROTATION_LENGTH)

        self.width(width)


if __name__ == '__main__':
    tree_level = 11  # choose
    tree = Tree_Fractal(tree_level)
    turtle.done()