297.二叉树的序列化和反序列化

难度：困难
题目描述：

思路总结：终于又来了一道困难题，还是一样的酸爽，不过这题可以思路很简单的进行解决。

题解一：
pop(0)比较费时间，用iter——next，可以达到不错的效果，具体代码就不粘了，结果给一下。

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Codec:

    def serialize(self, root):
        """Encodes a tree to a single string.
        
        :type root: TreeNode
        :rtype: str
        """
        def predfs(node, res):
            if not node:
                res += "None,"
            else:
                res += str(node.val) +","
                res = predfs(node.left, res)
                res = predfs(node.right, res)
            return res
        return predfs(root, "")

    def deserialize(self, data):
        """Decodes your encoded data to tree.
        
        :type data: str
        :rtype: TreeNode
        """
        def helper(l):
            if l[0] == "None":
                l.pop(0)
                return None
            cur = TreeNode(l.pop(0))
            cur.left = helper(l)
            cur.right = helper(l)
            return cur
        data = data.split(",")
        return helper(data)
#iter改进后的helper函数
		def helper():
            tmp = next(l)
            if tmp == "None":
                return None
            cur = TreeNode(tmp)
            cur.left = helper()
            cur.right = helper()
            return cur

# Your Codec object will be instantiated and called as such:
# codec = Codec()
# codec.deserialize(codec.serialize(root))

题解一结果：

改进后：空间使用增大，时间减半。