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原來 中序加前序去重建tree是有限制的，val不能重複!!

Problem

Serialization is the process of 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 tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.

Clarification: The input/output format is the same as how LeetCode serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.

 

Example 1:

Input: root = [1,2,3,null,null,4,5]Output: [1,2,3,null,null,4,5]

Example 2:

Input: root = []Output: []

Example 3:

Input: root = [1]Output: [1]

Example 4:

Input: root = [1,2]Output: [1,2]

 

Constraints:

  • The number of nodes in the tree is in the range [0, 104].
  • -1000 <= Node.val <= 1000

Sol: dfs

不論哪一種走訪，都是看序列化時怎麼走，反序列化就照著走

class Codec:
    def inorder(self,r):
        if not r:
            self.ino.append('x')
        else:
            self.ino.append(str(r.val))
            self.inorder(r.left)
            self.inorder(r.right)
    def serialize(self, root):
        if not root:
            return ""
        else:
            self.ino = []
            self.inorder(root)
            #print(self.ino)
            return ','.join(self.ino)
    def dfs(self):
        self.i += 1
        if self.data[self.i] != 'x':
            root = TreeNode()
            root.val = int(self.data[self.i])
            root.left = self.dfs()
            root.right = self.dfs()
            return root
        else:
            return None
    def deserialize(self, data):
        if not data:
            return None
        self.data = data.split(',')
        self.i = -1
        
        return self.dfs()

Sol: bfs

class Codec:
    def serialize(self, root):
        if not root:
            return ""
        else:
            ret = []
            q = deque([root])
            while q:
                tmp = q.popleft()
                if tmp:
                    ret.append(str(tmp.val))
                    q.append(tmp.left)
                    q.append(tmp.right)
                else:
                    ret.append('x')
            return ','.join(ret)

    def deserialize(self, data):
        if not data:
            return None
        else:
            data = data.split(',')
            ret = TreeNode()
            ret.val = int(data[0])
            q = deque([ret])
            i = 1
            while q and i < len(data):
                root = q.popleft()
                now = data[i] if i < len(data) else 'x'
                if now != 'x':
                    root.left = TreeNode()
                    root.left.val = int(now)
                    q.append(root.left)
                i += 1
                
                now = data[i] if i < len(data) else 'x'
                if now != 'x':
                    root.right = TreeNode()
                    root.right.val = int(now)
                    q.append(root.right)
                i += 1

        return ret