這是我的程式碼(它是不正確的,對的請看力扣,我這裡只是記錄一個有趣的函式)

import math


class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

    def __str__(self):
        return self.val


root = TreeNode(3)
t1 = TreeNode(9)
t2 = TreeNode(20)
t3 = TreeNode(None)
t4 = TreeNode(None)
t5 = TreeNode(15)
t6 = TreeNode(7)

t1.left = t3
t1.right = t4

t2.left = t5
t2.right = t6

root.left = t1
root.right = t2


class Solution(object):
    def zigzagLevelOrder(self, root):
        # 在寫這個的時候我以為root是滿二叉樹,也就是說沒有地方用None補齊
        node_list = self.ground(root)  # 這個函式是一個廣度遍歷二叉樹的函式

        lenght = len(node_list) + 1  # 滿二叉樹的節點等於2^n -1 ,其中n是二叉樹的層數
        cengji = int(math.log(lenght, 2))   # 這個是我用對數函式求出來的二叉樹層數
        k = 0  # 其中k表示每次取擷取node_list的起始位置
        ret_list = []
        for i in range(1, cengji + 1):  
            num = 2 ** (i-1)   # 計算一層有幾個節點
            cut_list = node_list[k: num + k]  # num + k是擷取的最後位置
            if i % 2:
                cut_list = cut_list
            else:
                cut_list = cut_list[::-1]

            cut_list = [item for item in cut_list if item]

            ret_list.append(cut_list)
            k += num
        return ret_list

    def ground(self, root):
        if not root:
            return
        retList = []
        queue = []
        queue.append(root)
        while queue:
            r = queue.pop(0)
            retList.append(r.val)
            if r.left:
                queue.append(r.left)

            if r.right:
                queue.append(r.right)

        return retList


s1 = Solution()
print(s1.zigzagLevelOrder(root))