剑指 Offer II 048. 序列化与反序列化二叉树

题目描述

序列化是将一个数据结构或者对象转换为连续的比特位的操作，进而可以将转换后的数据存储在一个文件或者内存中，同时也可以通过网络传输到另一个计算机环境，采取相反方式重构得到原数据。

请设计一个算法来实现二叉树的序列化与反序列化。这里不限定你的序列 / 反序列化算法执行逻辑，只需要保证一个二叉树可以被序列化为一个字符串并且将这个字符串反序列化为原始的树结构。

示例 1：

输入：root = [1,2,3,null,null,4,5]
输出：[1,2,3,null,null,4,5]

示例 2：

输入：root = []
输出：[]

示例 3：

输入：root = [1]
输出：[1]

示例 4：

输入：root = [1,2]
输出：[1,2]

提示：

输入输出格式与 LeetCode 目前使用的方式一致，详情请参阅 LeetCode 序列化二叉树的格式。你并非必须采取这种方式，也可以采用其他的方法解决这个问题。
树中结点数在范围 [0, 10⁴] 内
-1000 <= Node.val <= 1000

注意：本题与主站 297 题相同：https://leetcode.cn/problems/serialize-and-deserialize-binary-tree/

解法

方法一

Python3JavaC++JavaScript

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Codec:
    def serialize(self, root):
        """Encodes a tree to a single string.

        :type root: TreeNode
        :rtype: str
        """

        if root is None:
            return ''
        res = []

        def preorder(root):
            if root is None:
                res.append("#,")
                return
            res.append(str(root.val) + ",")
            preorder(root.left)
            preorder(root.right)

        preorder(root)
        return ''.join(res)

    def deserialize(self, data):
        """Decodes your encoded data to tree.

        :type data: str
        :rtype: TreeNode
        """
        if not data:
            return None
        vals = data.split(',')

        def inner():
            first = vals.pop(0)
            if first == '#':
                return None
            return TreeNode(int(first), inner(), inner())

        return inner()


# Your Codec object will be instantiated and called as such:
# ser = Codec()
# deser = Codec()
# ans = deser.deserialize(ser.serialize(root))

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Codec {
    private static final String NULL = "#";
    private static final String SEP = ",";

    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        if (root == null) {
            return "";
        }
        StringBuilder sb = new StringBuilder();
        preorder(root, sb);
        return sb.toString();
    }

    private void preorder(TreeNode root, StringBuilder sb) {
        if (root == null) {
            sb.append(NULL + SEP);
            return;
        }
        sb.append(root.val + SEP);
        preorder(root.left, sb);
        preorder(root.right, sb);
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if (data == null || "".equals(data)) {
            return null;
        }
        List<String> vals = new LinkedList<>();
        for (String x : data.split(SEP)) {
            vals.add(x);
        }
        return deserialize(vals);
    }

    private TreeNode deserialize(List<String> vals) {
        String first = vals.remove(0);
        if (NULL.equals(first)) {
            return null;
        }
        TreeNode root = new TreeNode(Integer.parseInt(first));
        root.left = deserialize(vals);
        root.right = deserialize(vals);
        return root;
    }
}

// Your Codec object will be instantiated and called as such:
// Codec ser = new Codec();
// Codec deser = new Codec();
// TreeNode ans = deser.deserialize(ser.serialize(root));

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Codec {
public:
    // Encodes a tree to a single string.
    string serialize(TreeNode* root) {
        if (!root) return "";
        string s = "";
        preorder(root, s);
        return s;
    }

    void preorder(TreeNode* root, string& s) {
        if (!root)
            s += "# ";
        else {
            s += to_string(root->val) + " ";
            preorder(root->left, s);
            preorder(root->right, s);
        }
    }

    // Decodes your encoded data to tree.
    TreeNode* deserialize(string data) {
        if (data == "") return nullptr;
        stringstream ss(data);
        return deserialize(ss);
    }

    TreeNode* deserialize(stringstream& ss) {
        string first;
        ss >> first;
        if (first == "#") return nullptr;
        TreeNode* root = new TreeNode(stoi(first));
        root->left = deserialize(ss);
        root->right = deserialize(ss);
        return root;
    }
};

// Your Codec object will be instantiated and called as such:
// Codec ser, deser;
// TreeNode* ans = deser.deserialize(ser.serialize(root));

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */

/**
 * Encodes a tree to a single string.
 *
 * @param {TreeNode} root
 * @return {string}
 */
var serialize = function (root) {
    return rserialize(root, '');
};

/**
 * Decodes your encoded data to tree.
 *
 * @param {string} data
 * @return {TreeNode}
 */
var deserialize = function (data) {
    const dataArray = data.split(',');
    return rdeserialize(dataArray);
};

const rserialize = (root, str) => {
    if (root === null) {
        str += '#,';
    } else {
        str += root.val + '' + ',';
        str = rserialize(root.left, str);
        str = rserialize(root.right, str);
    }
    return str;
};

const rdeserialize = dataList => {
    if (dataList[0] === '#') {
        dataList.shift();
        return null;
    }

    const root = new TreeNode(parseInt(dataList[0]));
    dataList.shift();
    root.left = rdeserialize(dataList);
    root.right = rdeserialize(dataList);

    return root;
};

/**
 * Your functions will be called as such:
 * deserialize(serialize(root));
 */

剑指 Offer II 048. 序列化与反序列化二叉树

题目描述

解法

方法一

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