Python实现的序列化和反序列化二叉树算法示例
本文实例讲述了Python实现的序列化和反序列化二叉树算法。分享给大家供大家参考，具体如下：
题目描述
请实现两个函数，分别用来序列化和反序列化二叉树
序列化二叉树
先序遍历二叉树
def recursionSerialize(self, root):
series = ''
if root == None:
series += ',$'
else:
series += (',' + str(root.val))
series += self.recursionSerialize(root.left)
series += self.recursionSerialize(root.right)
return series
def Serialize(self, root):
return self.recursionSerialize(root)[1:]
结果：
root = TreeNode(11) root.left = TreeNode(2) root.right = TreeNode(3) series = Solution().Serialize(root) print(series) >>>11,2,$,$,3,$,$
反序列化
先构建根节点，然后左节点，右节点，同样是递归
注意由于使用的是字符串的表示形式，可以先转化为list，
print(series.split(','))
>>>['11', '2', '$', '$', '3', '$', '$']
然后再处理就不需要将大于10的数字转换过来了：
def getValue(self, s, sIndex): #处理超过10的数字，将数字字符转变为数字
val = 0
while ord(s[sIndex]) <= ord('9') and ord(s[sIndex]) >= ord('0'):
val = val * 10 + int(s[sIndex])
sIndex += 1
return val, sIndex - 1
下面是反序列化的递归函数：
def Deserialize(self, s):
if self.sIndex < len(s):
if s[self.sIndex] == ',':
self.sIndex += 1
if s[self.sIndex] == '$':
return None
val, self.sIndex = self.getValue(s, self.sIndex)
treeNode = TreeNode(val)
self.sIndex += 1
treeNode.left = self.Deserialize(s)
self.sIndex += 1
treeNode.right = self.Deserialize(s)
return treeNode
完整解法
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution:
def __init__(self):
self.sIndex = 0
def recursionSerialize(self, root):
series = ''
if root == None:
series += ',$'
else:
series += (',' + str(root.val))
series += self.recursionSerialize(root.left)
series += self.recursionSerialize(root.right)
return series
def Serialize(self, root):
return self.recursionSerialize(root)[1:]
def getValue(self, s, sIndex): #处理超过10的数字，将数字字符转变为数字
val = 0
while ord(s[sIndex]) <= ord('9') and ord(s[sIndex]) >= ord('0'):
val = val * 10 + int(s[sIndex])
sIndex += 1
return val, sIndex - 1
def Deserialize(self, s):
if self.sIndex < len(s):
if s[self.sIndex] == ',':
self.sIndex += 1
if s[self.sIndex] == '$':
return None
val, self.sIndex = self.getValue(s, self.sIndex)
treeNode = TreeNode(val)
self.sIndex += 1
treeNode.left = self.Deserialize(s)
self.sIndex += 1
treeNode.right = self.Deserialize(s)
return treeNode
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