449. Serialize and Deserialize BST

题目描述

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 search tree. There is no restriction on how your serialization/deserialization algorithm should work. You just 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.

Note: Do not use class member/global/static variables to store states. Your serialize and deserialize algorithms should be stateless.

题目链接

https://leetcode.com/problems/serialize-and-deserialize-bst/

方法思路

public class Codec {
    //Runtime: 6 ms, faster than 84.94%
    //Memory Usage: 40.1 MB, less than 43.69%
    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        if(root == null){
            return null;
        }
        StringBuffer sb = new StringBuffer();
        serializeCore(root, sb);
        return sb.toString();
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if(data == null){
            return null;
        }
        String[] nodes = data.split(",");
        TreeNode res = deserializeCore(nodes, 0, nodes.length - 1);
        return res;
    }
    
    //前序序列化
    private void serializeCore(TreeNode root, StringBuffer sb){
        if(root == null){
            return;
        }
        sb.append(root.val);
        sb.append(",");
        serializeCore(root.left, sb);
        serializeCore(root.right, sb);
    }
    
    //前序反序列化
    private TreeNode deserializeCore(String[] nodes, int start, int end){
        if(start > end){
            return null;
        }
        if(start == end){
            return new TreeNode(Integer.parseInt(nodes[start]));
        }
        TreeNode root = new TreeNode(Integer.parseInt(nodes[start]));
        int index = start + 1;
        while(index <= end && Integer.parseInt(nodes[index]) < Integer.parseInt(nodes[start])){
            ++index;
        }
        root.left = deserializeCore(nodes, start + 1, index - 1);
        root.right = deserializeCore(nodes, index, end);
        return root;
    }
}