Skip to content
New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

[LeetCode] 297. Serialize and Deserialize Binary Tree #297

Open
grandyang opened this issue May 30, 2019 · 0 comments
Open

[LeetCode] 297. Serialize and Deserialize Binary Tree #297

grandyang opened this issue May 30, 2019 · 0 comments

Comments

@grandyang
Copy link
Owner

grandyang commented May 30, 2019

 

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;
    }
};

 

Github 同步地址:

#297

 

参考资料:

https://leetcode.com/problems/serialize-and-deserialize-binary-tree/

https://leetcode.com/problems/serialize-and-deserialize-binary-tree/discuss/74253/Easy-to-understand-Java-Solution

https://leetcode.com/problems/serialize-and-deserialize-binary-tree/discuss/74261/Easy-to-understand-java-solution

https://leetcode.com/problems/serialize-and-deserialize-binary-tree/discuss/74264/Short-and-straight-forward-BFS-Java-code-with-a-queue

 

LeetCode All in One 题目讲解汇总(持续更新中...)

Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment
Labels
None yet
Projects
None yet
Development

No branches or pull requests

1 participant