LeetCode|【LeetCode】297. Serialize and Deserialize Binary Tree 解题报告（Python） LeetCode|算法

【LeetCode】297. Serialize and Deserialize Binary Tree 解题报告（Python） 【LeetCode|【LeetCode】297. Serialize and Deserialize Binary Tree 解题报告（Python）】标签： LeetCode
题目地址：https://leetcode.com/problems/serialize-and-deserialize-binary-tree/description/
题目描述： 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 / \ 23 / \ 45

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.
题目大意序列化，解序列化一棵二叉树。
解题方法和449. Serialize and Deserialize BST多么的像呀！之前我说，只知道前序遍历是没法确定一个树的，我说的不严谨。如果前序遍历的过程中记录下哪些位置是空节点的话，就是可以确定这棵树的。LeetCode的官方树的构建就是这样的。
因此，我们采用和上题同样的方法，只不过需要把空节点记录下来。然后在反序列化时把它再变成空节点即可。

# 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 """ vals = [] def preOrder(root): if not root: vals.append('#') else: vals.append(str(root.val)) preOrder(root.left) preOrder(root.right) preOrder(root) return ' '.join(vals)def deserialize(self, data): """Decodes your encoded data to tree.:type data: str :rtype: TreeNode """ vals = collections.deque(val for val in data.split()) def build(): if vals: val = vals.popleft() if val == '#': return None root = TreeNode(int(val)) root.left = build() root.right = build() return root return build()# Your Codec object will be instantiated and called as such: # codec = Codec() # codec.deserialize(codec.serialize(root))

日期 2018 年 3 月 15 日 –雾霾消散，春光明媚