这个题解使用层序遍历(广度优先搜索)来序列化二叉树。通过使用队列来迭代访问每个节点,并将其值存储为字符串。对于空节点,它使用'null'来表示。反序列化过程中,同样采用层序遍历的方法。首先将字符串分割成节点数组,然后通过迭代构建每个节点的左右子节点。这种方法直观且符合大多数人对树结构的遍历习惯。
时间复杂度: O(n)
空间复杂度: O(n)
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Codec:
def serialize(self, root):
"""Encodes a tree to a single string.
:type root: TreeNode
:rtype: str
"""
if root is None:
return '[]'
res = []
queue = deque()
queue.append(root)
while queue:
node = queue.popleft()
if node:
res.append(str(node.val))
queue.append(node.left)
queue.append(node.right)
else:
res.append("null")
# Remove unnecessary nulls from end to optimize the result
while res[-1] == "null":
res.pop()
return '[' + ','.join(res) + ']'
def deserialize(self, data):
"""Decodes your encoded data to tree.
:type data: str
:rtype: TreeNode
"""
if data == '[]':
return
vals = data[1:-1].split(',')
root = TreeNode(int(vals[0]))
queue = deque()
queue.append(root)
i = 1
while i < len(vals):
node = queue.popleft()
if vals[i] != 'null':
node.left = TreeNode(int(vals[i]))
queue.append(node.left)
i += 1
if i < len(vals) and vals[i] != 'null':
node.right = TreeNode(int(vals[i]))
queue.append(node.right)
i += 1
return root
# Your Codec object will be instantiated and called as such:
# codec = Codec()
# codec.deserialize(codec.serialize(root))