# LeetCode: 449. Serialize and Deserialize BST

## 题目¶

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.

## 解法¶

• [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

• 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，

```# 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
```