LeetCode in Kotlin
449. Serialize and Deserialize BST
Medium
Serialization is 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 search tree. There is no restriction on how your serialization/deserialization algorithm should work. You need to ensure that a binary search tree can be serialized to a string, and this string can be deserialized to the original tree structure.
The encoded string should be as compact as possible.
Example 1:
Input: root = [2,1,3]
Output: [2,1,3]
Example 2:
Input: root = []
Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 104]. 0 <= Node.val <= 104- The input tree is guaranteed to be a binary search tree.
Solution
import com_github_leetcode.TreeNode
/*
* Definition for a binary tree node.
* class TreeNode(var `val`: Int) {
* var left: TreeNode? = null
* var right: TreeNode? = null
* }
*/
class Codec {
private var cur = 0
// Encodes a tree to a single string.
fun serialize(root: TreeNode?): String {
val sb = StringBuilder()
serializeDfs(root, sb)
return sb.toString()
}
private fun serializeDfs(root: TreeNode?, sb: StringBuilder) {
if (root == null) {
sb.append(SPLIT)
return
}
sb.append((root.`val` + MIN).toChar())
serializeDfs(root.left, sb)
serializeDfs(root.right, sb)
}
// Decodes your encoded data to tree.
fun deserialize(data: String): TreeNode? {
cur = 0
return deserializeDFS(data.toCharArray())
}
private fun deserializeDFS(data: CharArray): TreeNode? {
if (data[cur] == SPLIT) {
cur++
return null
}
val node = TreeNode(data[cur++].code - MIN)
node.left = deserializeDFS(data)
node.right = deserializeDFS(data)
return node
}
companion object {
private const val SPLIT = 0.toChar()
private const val MIN = 1
}
}
/*
* Your Codec object will be instantiated and called as such:
* val ser = Codec()
* val deser = Codec()
* val tree: String = ser.serialize(root)
* val ans = deser.deserialize(tree)
* return ans
*/