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:
# 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
"""
# corner case
if root is None:
return ""
# regular case
out = [str(root.val)] # append value when enqueue
q = deque([root]) # put root in queue
while q: # while queue not empty
# pop
node = q.popleft()
# children
for attr in ['left', 'right']:
if getattr(node, attr) is None:
# corner case
out.append('*')
else:
# regular case
## Process the child
out.append(str(getattr(node, attr).val)) # append value when enqueue
## append the child
q.append(getattr(node, attr))
return ','.join(out) # list to string
def deserialize(self, data):
"""Decodes your encoded data to tree.
:type data: str
:rtype: TreeNode
"""
# corner case
if data == "":
return None
# regular case
data = data.split(',') # string to list
root = TreeNode(int(data[0])) # create node when enqueue
q = deque([root]) # put root in queue
i = 1 # index of data
while q: # while queue not empty
# pop
node = q.popleft()
# children
for attr in ['left', 'right']:
# corner case dealt with default value
if data[i] != '*':
# regular case
## Process
setattr(node, attr, TreeNode(int(data[i]))) # create node when enqueue
## append the child
q.append(getattr(node, attr))
i += 1 # move data index forward
return root # return root
# Your Codec object will be instantiated and called as such:
# ser = Codec()
# deser = Codec()
# ans = deser.deserialize(ser.serialize(root))
思路
while q:
node = q.popleft()
node = TreeNode(int(data[i]))
for attr in ['left', 'right']:
q.append(getattr(node, attr))
node = TreeNode(int(data[i]))
这一步,建立了新的变量 node
。
所以必须在 母节点 上对 子树 赋值,
而不能将子树 压入队列 之后再赋值。python 细节
data = [int(d) for d in data.split(',')]
不行,是因为中间有*
getattr
和 setattr
不是成员函数,而是 built-indef deserialize(self, data):
"""Decodes your encoded data to tree.
:type data: str
:rtype: TreeNode
"""
if data == "":
return None
data = deque(data.split(','))
root = TreeNode(int(data.popleft()))
q = deque([root])
while q:
node = q.popleft()
for attr in ['left', 'right']:
val = data.popleft()
if val != '*':
setattr(node, attr, TreeNode(int(val)))
q.append(getattr(node, attr))
return root
1,2,3,4,5,6,7,8,9,10,11,12,13,14,37,38,39,40,41,42,66,67,68,69
20
19
21
22
24
31
33
23
25
26,27
29,30
32
34
28
35
18
15,16
17
47
48
49
50
51
52
54
59
61
53
55
56
57,58
60
62
63
64
46
43,44
45