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Copy path297 Serialize and Deserialize Binary Tree.py
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297 Serialize and Deserialize Binary Tree.py
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"""
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.
Author: Rajeev Ranjan
"""
from collections import deque
class TreeNode(object):
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Codec:
def serialize(self, root):
"""
bfs
Encodes a tree to a single string.
encode to: 1, 2, 3, null, null, 4, 5, null, null, null, null
:type root: TreeNode
:rtype: str
"""
if not root:
return "null"
ret = []
q = []
q.append(root)
ret.append(str(root.val)) # add result when enqueue
while q:
l = len(q)
for i in xrange(l):
cur = q[i]
if cur.left: q.append(cur.left)
ret.append(self.encode(cur.left))
if cur.right: q.append(cur.right)
ret.append(self.encode(cur.right))
q = q[l:]
return ",".join(ret)
def deserialize(self, data):
"""
Decodes your encoded data to tree.
decode: 1, 2, 3, null, null, 4, 5, null, null, null, null
:type data: str
:rtype: TreeNode
"""
lst = data.split(",")
root = self.decode(lst[0])
q = deque([root])
i = 1
while q:
cur = q.popleft()
if i < len(lst):
cur.left = self.decode(lst[i])
i += 1
if cur.left: q.append(cur.left)
if i < len(lst):
cur.right = self.decode(lst[i])
i += 1
if cur.right: q.append(cur.right)
return root
def decode(self, s):
if s == "null":
return None
else:
return TreeNode(int(s))
def encode(self, node):
if not node:
return "null"
else:
return str(node.val)