Binary Tree Serialization
Question
- lintcode: (7) Binary Tree Serialization
Design an algorithm and write code to serialize and deserialize a binary tree.
Writing the tree to a file is called 'serialization'
and reading back from the file to reconstruct
the exact same binary tree is 'deserialization'.
There is no limit of how you deserialize or serialize a binary tree,
you only need to make sure you can serialize a binary tree to a string
and deserialize this string to the original structure.
Have you met this question in a real interview? Yes
Example
An example of testdata: Binary tree {3,9,20,#,#,15,7},
denote the following structure:
3
/ \
9 20
/ \
15 7
Our data serialization use bfs traversal.
This is just for when you got wrong answer and want to debug the input.
You can use other method to do serializaiton and deserialization.
题解
根据之前由前序，中序，后序遍历恢复二叉树的经验，确定根节点的位置十分重要（但是这里可能有重复元素，故和之前的题目不太一样）。能直接确定根节点的有前序遍历和广度优先搜索，其中较为简洁的为前序遍历。序列化较为简单，但是反序列化的实现不太容易。需要借助字符串解析工具。
Python
"""
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""
"""
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""
class Solution0:
'''
@param root: An object of TreeNode, denote the root of the binary tree.
This method will be invoked first, you should design your own algorithm
to serialize a binary tree which denote by a root node to a string which
can be easily deserialized by your own "deserialize" method later.
'''
def serialize(self, root):
if not root:
return ''
def post_order(root):
if root:
post_order(root.left)
post_order(root.right)
ret[0] += str(root.val) + ','
else:
ret[0] += '#,'
ret = ['']
post_order(root)
return ret[0][:-1] # remove last ,
'''
@param data: A string serialized by your serialize method.
This method will be invoked second, the argument data is what exactly
you serialized at method "serialize", that means the data is not given by
system, it's given by your own serialize method. So the format of data is
designed by yourself, and deserialize it here as you serialize it in
"serialize" method.
'''
def deserialize(self, data):
if not data:
return
nodes = data.split(',')
def post_order(nodes):
if nodes[-1] == '#':
nodes.pop()
return None
root = TreeNode(int(nodes.pop()))
root.right = post_order(nodes)
root.left = post_order(nodes)
return root
return post_order(nodes)
class Solution1:
'''
@param root: An object of TreeNode, denote the root of the binary tree.
This method will be invoked first, you should design your own algorithm
to serialize a binary tree which denote by a root node to a string which
can be easily deserialized by your own "deserialize" method later.
'''
def serialize(self, root):
if not root:
return ''
def pre_order(root):
if root:
ret[0] += str(root.val) + ','
pre_order(root.left)
pre_order(root.right)
else:
ret[0] += '#,'
ret = ['']
pre_order(root)
return ret[0][:-1] # remove last ,
'''
@param data: A string serialized by your serialize method.
This method will be invoked second, the argument data is what exactly
you serialized at method "serialize", that means the data is not given by
system, it's given by your own serialize method. So the format of data is
designed by yourself, and deserialize it here as you serialize it in
"serialize" method.
'''
def deserialize(self, data):
if not data:
return
nodes = data.split(',')
self.i = 0
def pre_order(nodes):
if nodes[self.i] == '#':
return None
root = TreeNode(int(nodes[self.i]))
self.i += 1
root.left = pre_order(nodes)
self.i += 1
root.right = pre_order(nodes)
return root
return pre_order(nodes)
import collections
class Solution2:
'''
@param root: An object of TreeNode, denote the root of the binary tree.
This method will be invoked first, you should design your own algorithm
to serialize a binary tree which denote by a root node to a string which
can be easily deserialized by your own "deserialize" method later.
'''
def serialize(self, root):
if not root:
return
ret = []
queue = collections.deque()
queue.append(root)
while queue:
node = queue.popleft()
if node:
queue.append(node.left)
queue.append(node.right)
ret.append(str(node.val))
else:
ret.append('#')
return ','.join(ret)
'''
@param data: A string serialized by your serialize method.
This method will be invoked second, the argument data is what exactly
you serialized at method "serialize", that means the data is not given by
system, it's given by your own serialize method. So the format of data is
designed by yourself, and deserialize it here as you serialize it in
"serialize" method.
'''
def deserialize(self, data):
if not data:
return
nodes = data.split(',')
root = TreeNode(int(nodes[0]))
i = 1
queue = collections.deque()
queue.append(root)
while queue:
node = queue.popleft()
if nodes[i] == '#':
node.left = None
else:
t = TreeNode(int(nodes[i]))
node.left = t
queue.append(t)
i += 1
if nodes[i] == '#':
node.right = None
else:
t = TreeNode(int(nodes[i]))
node.right = t
queue.append(t)
i += 1
return root
源码分析
第零种解法是后序遍历（推荐）， 在serialize的时候， 需要先左->右->中。 在deserialize的时候，因为是从最后一个值开始pop， 构成tree的时候， 就应该先中->右->左。
第一种解法是前序遍历， 其中巧妙的利用了python的closure， 在serialize中， 利用了list mutable 的特性， 修改了ret中的值。 deserialize中， 利用了self.i来储存instance variable。
第二种解法是广度遍历。 在deserialize的时候， 保持一个index i，记录用过的node。
"""
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
this.val = val
this.left, this.right = None, None
"""
class Solution:
'''
@param root: An object of TreeNode, denote the root of the binary tree.
This method will be invoked first, you should design your own algorithm
to serialize a binary tree which denote by a root node to a string which
can be easily deserialized by your own "deserialize" method later.
'''
def serialize(self, root):
return TreeNode.toString(root)
'''
@param data: A string serialized by your serialize method.
This method will be invoked second, the argument data is what exactly
you serialized at method "serialize", that means the data is not given by
system, it's given by your own serialize method. So the format of data is
designed by yourself, and deserialize it here as you serialize it in
"serialize" method.
'''
def deserialize(self, data):
return TreeNode.parse(data)