LeetCode 297. Serialize and Deserialize Binary Tree

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.

Example: 

You may serialize the following tree:

    1
   / \
  2   3
     / \
    4   5

as "[1,2,3,null,null,4,5]"

Clarification: The above 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.

Note: Do not use class member/global/static variables to store states. Your serialize and deserialize algorithms should be stateless.

解析：对二叉树进行序列化和反序列化。

序列化的过程类似于二叉树的层次遍历的过程，考虑BFS借助队列进行。先将树的跟节点入队列，然后打印数值，然后再把根节点的左子树和右子树分别；进队列，然后循环出队列遍历。可以得到序列化的部分代码如下：

    // Encodes a tree to a single string.
    string serialize(TreeNode* root) {
        ostringstream out;
        queue<TreeNode*> que;
        if(root)
            que.push(root);
        while(!que.empty())
        {
            TreeNode* t = que.front();
            que.pop();
            if(t)
            {
                out << t->val <<' ';
                que.push(t->left);
                que.push(t->right);
            }else
            {
                out <<"# ";
            }
        }
        return out.str();
    }

对于反序列化，也是采用队列，进行DFS处理。先取出第一个字符，生成根节点，然后把根节点入队列。然后分别取出后面的字符，分别生成二叉树节点，作为根节点的左子树和右子树。

反序列化的代码如下：

// Decodes your encoded data to tree.
    TreeNode* deserialize(string data) {
        if(data.empty())
            return NULL;
        istringstream in(data);
        queue<TreeNode*> que;
        string val;
        in >> val;
        TreeNode *root = new TreeNode(stoi(val)), *cur = root;
        que.push(cur);
        while(!que.empty())
        {
            TreeNode *t = que.front();
            que.pop();
            if(!(in >> val))
                break;
            if(val != "#")
            {
                cur = new TreeNode(stoi(val));
                que.push(cur);
                t->left = cur;
            }
            if(!(in >> val))
                break;
            if(val != "#")
            {
                cur = new TreeNode(stoi(val));
                que.push(cur);
                t->right = cur;
            }
        }
        return root;
    }