LeetCode: 297. Serialize and Deserialize Binary Tree

题目¶

原题地址：https://leetcode.com/problems/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.

Clarification: The input/output format is the same as how LeetCode serializes a binary tree . You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.

Example 1:

Input: root = [1,2,3,null,null,4,5]
Output: [1,2,3,null,null,4,5]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

Example 4:

Input: root = [1,2]
Output: [1,2]

Constraints:

• The number of nodes in the tree is in the range [0, 10^4].
• -1000 <= Node.val <= 1000

题目大意是，设计一个类实现序列化和反序列化一个二叉树的功能

解法¶

序列化，中序遍历将节点的值用空格分隔组成一个字符串，通过使用 N 标识空节点:

• [1,null,2] 将序列化为 1 N 2 N N
• [2,1,3] 将序列化为 2 1 N N 3 N N
• [5,3,6,2,4,null,7] 将序列化为 5 3 2 N N 4 N N 6 N 7 N N

反序列化，按空格读取字符串中包含的所有节点的值，然后基于读取处理的值列表重建二叉树： * 按照中序遍历的过程来重建二叉树 * 如果当前值是 N 说明是空节点 * 否则就是普通节点

这个思路的 Python 代码类似下面这样：

# 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."""
        if root is None:
            return 'N'

        val = '{}'.format(root.val)
        left = self.serialize(root.left)
        right = self.serialize(root.right)

        return '{} {} {}'.format(val, left, right)


    def deserialize(self, data):
        """Decodes your encoded data to tree."""
        values = data.split()
        root = self._build_tree(values)

        return root

    def _build_tree(self, values):
        if not values:
            return None

        next_value = values.pop(0)
        if next_value == 'N':
            return None

        val = int(next_value)
        root = TreeNode(val)
        root.left = self._build_tree(values)
        root.right = self._build_tree(values)

        return root


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