Serialize and Deserialize Binary Tree

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

For example, you may serialize the following tree

    1
   / \
  2   3
     / \
    4   5

as"[1,2,3,null,null,4,5]", just the same as how LeetCode OJ serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.

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

这道题让我们对二叉树进行序列化和去序列化的操作。序列化就是将一个数据结构或物体转化为一个位序列，可以存进一个文件或者内存缓冲器中，然后通过网络连接在相同的或者另一个电脑环境中被还原，还原的过程叫做去序列化。现在让我们来序列化和去序列化一个二叉树，并给了我们例子。这题有两种解法，分别为先序遍历的递归解法和层序遍历的非递归解法。先来看先序遍历的递归解法，非常的简单易懂，我们需要接入输入和输出字符串流istringstream和ostringstream，对于序列化，我们从根节点开始，如果节点存在，则将值存入输出字符串流，然后分别对其左右子节点递归调用序列化函数即可。对于去序列化，我们先读入第一个字符，以此生成一个根节点，然后再对根节点的左右子节点递归调用去序列化函数即可

解法I：二叉树的先序遍历

# 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 doit(node):
            if node:
                vals.append(str(node.val))
                doit(node.left)
                doit(node.right)
            else:
                vals.append('#')
        vals = []
        doit(root)
        return ' '.join(vals)

    def deserialize(self, data):
        def doit():
            val = next(vals)
            if val == '#':
                return None
            node = TreeNode(int(val))
            node.left = doit()
            node.right = doit()
            return node
        vals = iter(data.split())
        return doit()

The iter() method creates an object which can be iterated one element at a time.

# list of vowels
vowels = ['a', 'e', 'i', 'o', 'u']

vowelsIter = iter(vowels)

# prints 'a'
print(next(vowelsIter))

# prints 'e'
print(next(vowelsIter))

# prints 'i'
print(next(vowelsIter))

# prints 'o'
print(next(vowelsIter))

# prints 'u'
print(next(vowelsIter))

解法II：层序遍历的非递归解法

借助queue来做，本质是BFS算法