题目如下：

Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.

Design an algorithm to serialize and deserialize a binary search tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary search tree can be serialized to a string and this string can be deserialized to the original tree structure.

The encoded string should be as compact as possible.

Note: Do not use class member/global/static variables to store states. Your serialize and deserialize algorithms should be stateless.

解题思路：我的序列化的方法是"根左右"的顺序进行遍历，每个节点的值之间以'#'分割。那么在反序列化的时候，首先把序列化的结果按'#'分割成数组，很显然数组的第一个元素是根节点，同时遍历数组，找到第一个比根节点值大的节点，这个节点左边的元素属于根节点的左子树，自身及右边的元素属于根节点的右子树。接下来分别对左右子树递归，直到所有节点的构造完成为止。

代码如下：

# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Codec:    def serialize(self, root):        """Encodes a tree to a single string.                :type root: TreeNode        :rtype: str        """        self.ser = ''        def recurisve(node):            if node == None:                return            self.ser += '#'            self.ser += str(node.val)            recurisve(node.left)            recurisve(node.right)        recurisve(root)        return self.ser[1:]            def deserialize(self, data):        """Decodes your encoded data to tree.                :type data: str        :rtype: TreeNode        """        if len(data) == 0:            return None        val_list = [int(i) for i in data.split('#')]        root = TreeNode(-1)        def build(node,path):            if len(path) == 0:                return            rv = path[0]            left = []            inx = 0            for i in range(1,len(path)):                if path[i] < rv:                    inx = i                    left.append(path[i])                else:                    break            right = path[inx+1:]            node.val = rv            if len(left) > 0:                node.left = TreeNode(-1)                build(node.left,left)            if len(right) > 0:                node.right = TreeNode(-1)                build(node.right, right)        build(root,val_list)        return root        # Your Codec object will be instantiated and called as such:# codec = Codec()# codec.deserialize(codec.serialize(root))