感觉写了一座💩💩💩💩屎山💩💩💩💩，供大家观摩留念。。。

代码核心思想是BFS遍历，使用{}作为不同节点的分割（节点可能出现10这样的两个字符的数据。） 然后一堆判断。 咦。。。好恶心🤮🤮🤮🤮🤮

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    
    def Serialize(self, root):
        if root == None:
            return ""
        stack = [root]
        valstr = ""
        while stack != None and len(stack) !=0:
            c = stack.pop(0)
            if c.val == -1:
                valstr += "{#}"
                continue
            else:
                valstr += '{'+str(c.val)+'}'
            if c.left != None:
                stack.append(c.left)
            if c.left == None:
                stack.append(TreeNode(-1))
            if c.right != None:
                stack.append(c.right)
            if c.right == None:
                stack.append(TreeNode(-1))
        return valstr
            
    def Deserialize(self, s):
        if s == "":
            return None
        tmpstr = ""
        i = 0
        while i < len(s[1:]) and s[i] != '}':
            i +=1
            tmpstr = s[1:i]
        startindex = len(tmpstr) +2
        pp = TreeNode(int(tmpstr))
        pc = pp
        step = 0
        stack = [pc]
        p = None
        tmpstr = ""
        i = startindex
        while i < len(s):
            if s[i] == '{':
                stridx = i
                while i < len(s[1:]) and s[i] != '}':
                    i +=1
                tmpstr = s[stridx+1:i]
                i +=1
            if tmpstr == "#":
                if (step) % 2 == 1:
                    stack.pop(0)
                step +=1
                continue
            p = stack[0]
            if (step) % 2 == 0:
                p.left = TreeNode(int(tmpstr))
                stack.append(p.left)
            else:
                stack.pop(0)
                p.right = TreeNode(int(tmpstr))
                stack.append(p.right)
            step +=1
        return pp