题目描述

blablabla

样例

blablabla

算法1

$O(n^2)$

blablabla

时间复杂度分析：blablabla

P 代码

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


#serialize()采用bfs，对当前二叉树搜索，遍历vector，将当前节点左右儿子依次存入vector，空节点需要删去。
#deserialize()首先切割字符串，然后用isLeft标记是当前是左右儿子，存为队列首节点的左右儿子。
class Solution:

    def serialize(self, root):
        """Encodes a tree to a single string.

        :type root: TreeNode
        :rtype: str
        """
        if not root:
            return []

        res = []
        q = [root]
        while q:
            tmp = q.pop(0)
            if not tmp:
                res.append('null')
            else:
                res.append(tmp.val)
                q.append(tmp.left)
                q.append(tmp.right)
        return res

    def deserialize(self, data):
        """Decodes your encoded data to tree.

        :type data: str
        :rtype: TreeNode
        """
        if len(data) == 0:
            return None
        if data[0] == 'null':
            return None

        root = TreeNode(data.pop(0))
        q = [root]
        isLeft = True
        while data:
            tmp = data.pop(0)
            if tmp != 'null':
                node = TreeNode(tmp)
                q.append(node)
                if isLeft:
                    q[0].left = node
                else:
                    q[0].right = node
            if not isLeft:
                q.pop(0)
            isLeft = not isLeft
        return root