[LeetCode] 297. Serialize and Deserialize Binary Tree 二叉树的序列化和去序列化

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

这道题让对二叉树进行序列化和去序列化的操作。序列化就是将一个数据结构或物体转化为一个位序列，可以存进一个文件或者内存缓冲器中，然后通过网络连接在相同的或者另一个电脑环境中被还原，还原的过程叫做去序列化。现在让序列化和去序列化一个二叉树，并给了例子。这题有两种解法，分别为先序遍历的递归解法和层序遍历的非递归解法。先来看先序遍历的递归解法，非常的简单易懂，需要接入输入和输出字符串流 istringstream 和 ostringstream，对于序列化，从根节点开始，如果节点存在，则将值存入输出字符串流，然后分别对其左右子节点递归调用序列化函数即可。对于去序列化，先读入第一个字符，以此生成一个根节点，然后再对根节点的左右子节点递归调用去序列化函数即可，参见代码如下：

解法一：

class Codec {
public:
    // Encodes a tree to a single string.
    string serialize(TreeNode* root) {
        ostringstream out;
        serialize(root, out);
        return out.str();
    }
    // Decodes your encoded data to tree.
    TreeNode* deserialize(string data) {
        istringstream in(data);
        return deserialize(in);
    }
private:
    void serialize(TreeNode *root, ostringstream &out) {
        if (root) {
            out << root->val << ' ';
            serialize(root->left, out);
            serialize(root->right, out);
        } else {
            out << "# ";
        }
    }
    TreeNode* deserialize(istringstream &in) {
        string val;
        in >> val;
        if (val == "#") return nullptr;
        TreeNode *root = new TreeNode(stoi(val));
        root->left = deserialize(in);
        root->right = deserialize(in);
        return root;
    }
};

另一种方法是层序遍历的非递归解法，这种方法略微复杂一些，需要借助 queue 来做，本质是 BFS 算法，也不是很难理解，就是 BFS 算法的常规套路稍作修改即可，参见代码如下：

解法二：

class Codec {
public:
    // Encodes a tree to a single string.
    string serialize(TreeNode* root) {
        ostringstream out;
        queue<TreeNode*> q;
        if (root) q.push(root);
        while (!q.empty()) {
            TreeNode *t = q.front(); q.pop();
            if (t) {
                out << t->val << ' ';
                q.push(t->left);
                q.push(t->right);
            } else {
                out << "# ";
            }
        }
        return out.str();
    }
    // Decodes your encoded data to tree.
    TreeNode* deserialize(string data) {
        if (data.empty()) return nullptr;
        istringstream in(data);
        queue<TreeNode*> q;
        string val;
        in >> val;
        TreeNode *res = new TreeNode(stoi(val)), *cur = res;
        q.push(cur);
        while (!q.empty()) {
            TreeNode *t = q.front(); q.pop();
            if (!(in >> val)) break;
            if (val != "#") {
                cur = new TreeNode(stoi(val));
                q.push(cur);
                t->left = cur;
            }
            if (!(in >> val)) break;
            if (val != "#") {
                cur = new TreeNode(stoi(val));
                q.push(cur);
                t->right = cur;
            }
        }
        return res;
    }
};