DFS

• idea：以传tuple的形式分割左右子树与根结点；

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

class Solution:

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

        :type root: TreeNode
        :rtype: str
        """
        if not root:
            return "."
        left = self.serialize(root.left)
        right = self.serialize(root.right)
        s = (left, right, root.val)
        return s

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

        :type data: str
        :rtype: TreeNode
        """
        if not data or data[-1] == '.': return None
        left, right, val = data
        root = TreeNode(val)
        root.left = self.deserialize(left)
        root.right = self.deserialize(right)
        return root

栈模拟

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

class Solution:

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

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

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