题目¶
原题地址：https://leetcode.com/problems/serialize-and-deserialize-bst/
Serialization is 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 search tree. There is no restriction on how your serialization/deserialization algorithm should work. You need to ensure that a binary search tree can be serialized to a string, and this string can be deserialized to the original tree structure.
The encoded string should be as compact as possible.
Example 1:
Input: root = [2,1,3] Output: [2,1,3]
Example 2:
Input: root = [] Output: []
Constraints:
- The number of nodes in the tree is in the range [0, 10^4].
- 0 <= Node.val <= 10^4
- The input tree is guaranteed to be a binary search tree.
题目大意是，设计一个类实现序列化和反序列化一个 BST 的功能
解法¶
序列化，中序遍历将节点的值用空格分隔组成一个字符串:
- [1,null,2] 将序列化为 1 2
- [2,1,3] 将序列化为 2 1 3
- [5,3,6,2,4,null,7] 将序列化为 5 3 2 4 6 7
反序列化，按空格读取字符串中包含的所有节点的值，然后基于读取处理的值列表重建 BST：
- 按照中序遍历的过程来重建 BST
- 因为没有一个标识位标明哪里是空节点，所以需要在构建 BST 的时候 判断当前值是否符合假设的节点位置，比如， 预期当前值是左子树的 root 节点值，但是实际上它的值比 root 节点的值大， 说明 root 节点其实没有左子树， 预期当前值是右子树的 root 节点值，但是实际上它的值比 root 节点的值小， 说明 root 节点其实没有右子树，
这个思路的 Python 代码类似下面这样：
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Codec:
def serialize(self, root):
if root is None:
return ''
val = '{}'.format(root.val)
if root.left is not None:
val = '{} {}'.format(val, self.serialize(root.left))
if root.right is not None:
val = '{} {}'.format(val, self.serialize(root.right))
return val
def deserialize(self, data):
"""Decodes your encoded data to tree.
"""
values = data.split()
root = self._build_bst(values, -1, 10**4 + 1)
return root
def _build_bst(self, values, should_gt, should_lt):
if not values:
return None
next_value = int(values[0])
# 不是预期的左侧节点或右侧节点，说明这个位置应该为空
if next_value <= should_gt or next_value >= should_lt:
return None
val = int(values.pop(0))
root = TreeNode(val)
# 左子树的值应当小于 root 节点的值
root.left = self._build_bst(values, should_gt, val)
# 右子树的值应当大于 root 节点的值
root.right = self._build_bst(values, val, should_lt)
return root
# Your Codec object will be instantiated and called as such:
# Your Codec object will be instantiated and called as such:
# ser = Codec()
# deser = Codec()
# tree = ser.serialize(root)
# ans = deser.deserialize(tree)
# return ans
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