# coding: utf-8 class Solution: ''' @param root: An object of TreeNode, denote the root of the binary tree. This method will be invoked first, you should design your own algorithm to serialize a binary tree which denote by a root node to a string which can be easily deserialized by your own "deserialize" method later. ''' def serialize(self, root): # write your code here # 分层遍历,子节点为空用#代替,最后清理尾部的所有#。 ret = [] if not root: return ret level = [root] while level: new_level = [] for node in level: ret.append(str(node.val) if node else '#') if node: new_level.append(node.left) new_level.append(node.right) level = new_level i = len(ret) - 1 while i >= 0: if ret[i] == '#': ret.pop() i -= 1 else: break return ret ''' @param data: A string serialized by your serialize method. This method will be invoked second, the argument data is what exactly you serialized at method "serialize", that means the data is not given by system, it's given by your own serialize method. So the format of data is designed by yourself, and deserialize it here as you serialize it in "serialize" method. ''' def deserialize(self, data): # write your code here if not data: return None root = TreeNode(data[0]) level = [root] i = 1 while i < len(data): new_level = [] for node in level: node.left = TreeNode(data[i]) if data[i] != '#' else None if node.left: new_level.append(node.left) i += 1 if i < len(data): node.right = TreeNode(data[i]) if data[i] != '#' else None if node.right: new_level.append(node.right) i += 1 level = new_level return root # medium: http://lintcode.com/zh-cn/problem/binary-tree-serialization/