[LeetCode] 297. Serialize and Deserialize Binary Tree（二叉树的序列化和反序列化）

• Difficulty: Hard

• Related Topics: Tree, Design

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

Description

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.
序列化是将一数据结构或对象转化为比特序列，使其能够存储于文件或内存缓冲区中，或通过网络连接传输，以便在相同或不同的电脑上还原的过程。

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.
设计一个算法用于二叉树的序列化和反序列化。不限制算法的实现方式，只要保证你的序列化算法能将二叉树转化为字符串，该字符串能够作为反序列化算法的输入并还原出原先的树结构即可。

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.
澄清：样例的输入/输出是 LeetCode 序列化二叉树的方式。你并不需要遵循这种方式，所以发挥你的想象力，创造另一种格式。

Examples

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, 1e4].
• -1000 <= Node.val <= 1000

Solution

我没有那么多创意，为了后面刷树相关的题目方便写 testcase，我这里使用的格式与官方格式基本接近，只有两处不同：

1. 没有首尾的方括号。

2. null 用 # 表示，而非 null。

例如，Example 1 的树用我的格式解析后的结果是：1,2,3,#,#,4,5,#,#,#,#。末尾多余的 # 不影响反序列化。

序列化和反序列化都采用中序遍历的形式，代码如下：

/**
 * Definition for a binary tree node.
 * class TreeNode(var `val`: Int) {
 *     var left: TreeNode? = null
 *     var right: TreeNode? = null
 * }
 */

class Codec() {
    fun serialize(root: TreeNode?): String {
        if (root == null) {
            return ""
        }
        val result = arrayListOf<String>()
        val queue: Queue<TreeNode?> = LinkedList()
        queue.offer(root)

        while (queue.isNotEmpty()) {
            val node = queue.poll()
            if (node == null) {
                result.add("#")
                continue
            } else {
                result.add("${node.`val`}")
            }
            queue.offer(node.left)
            queue.offer(node.right)
        }

        return result.joinToString(separator = ",")
    }

    fun deserialize(data: String): TreeNode? {
        if (data.isBlank()) {
            return null
        }
        val valueQueue: Queue<String> = A(data.split(','))
        if (valueQueue.peek() == "#") {
            return null
        }
        val root = TreeNode(valueQueue.poll().toInt())

        val nodeQueue: Queue<TreeNode> = ArrayDeque()
        nodeQueue.offer(root)

        while (valueQueue.isNotEmpty()) {
            val curNode = nodeQueue.poll()

            val leftValue = valueQueue.poll()
            if (leftValue != "#") {
                val left = TreeNode(leftValue.toInt())
                curNode.left = left
                nodeQueue.offer(left)
            }

            val rightValue = valueQueue.poll()
            if (rightValue != "#") {
                val right = TreeNode(rightValue.toInt())
                curNode.right = right
                nodeQueue.offer(right)
            }
        }

        return root
    }
}

/**
 * Your Codec object will be instantiated and called as such:
 * var ser = Codec()
 * var deser = Codec()
 * var data = ser.serialize(longUrl)
 * var ans = deser.deserialize(data)
 */