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] == '$': return None val, self.sIndex = self.getValue(s, self.sIndex) treeNode = TreeNode(val) self.sIndex += 1 treeNode.left = self.Deserialize(s) self.sIndex += 1 treeNode.right = self.Deserialize(s) return treeNode
完整解法
class TreeNode: def __init__(self, x): self.val = x self.left = None self.right = Noneclass Solution: def __init__(self): self.sIndex = 0 def recursionSerialize(self, root): series = '' if root == None: series += ',$' 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 希望本文所述对大家Python程序设计有所帮助。
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