LeetCode 449. Serialize and Deserialize BST

注：题目及解法均来自于LeetCode

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.

AC代码1

常规解法，serialize的时候保存的是BST的先序遍历，节点与节点之间用”,”隔开。
deserialize的时候根据先序遍历的性质，serialize后的结果的第一位即是根节点，比根节点小的序列是根节点的左子树，比根节点大的是根节点的右子树，递归地找到左子树和右子树的根节点即可构造整棵树。

public class _449_Se_Derialize {
    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */

     static final String SEP = ",";

        // Encodes a tree to a single string.
        public String serialize(TreeNode root) {
            StringBuilder stringBuilder = new StringBuilder();
            if(root == null) return null;
            Stack stack = new Stack<>();
            stack.add(root);
            while(!stack.isEmpty()){
                root = stack.pop();
                stringBuilder.append(root.val).append(SEP);
                if(root.right != null) stack.push(root.right);//因为stack先进后出的性质，要先存储右节点，在存储左节点，这样先输出的是左节点。
                if(root.left != null) stack.push(root.left);
            }
            return stringBuilder.toString();
        }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if(data == null) return null;
        String[] strs = data.split(SEP);
        Queue queue = new LinkedList<>();
        for(String s : strs) queue.offer(Integer.parseInt(s));
        return getNode(queue);
    }

    private TreeNode getNode(Queue queue){
        if(queue.isEmpty()) return null;
        TreeNode root = new TreeNode(queue.poll());
        Queue smallQueue = new LinkedList<>();
        while(!queue.isEmpty() && queue.peek() < root.val)
            smallQueue.offer(queue.poll());
        root.left = getNode(smallQueue);
        root.right =  getNode(queue);
        return root;
    }

// Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.deserialize(codec.serialize(root));
}

AC代码2

思路与解法1相同，只不过更为精简，没有用到 stack 和 queue 的数据结构。

public class Codec {
    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        if (root==null) return "";
        StringBuilder sb = new StringBuilder();
        seriHelper(root, sb);
        return sb.toString();
    }
    private void seriHelper(TreeNode root, StringBuilder sb){
        if (root==null) return;
        sb.append((char)root.val);
        seriHelper(root.left, sb);
        seriHelper(root.right, sb);
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        return deseHelper(data.toCharArray(), Integer.MIN_VALUE, Integer.MAX_VALUE, new int[1]);
    }

    private TreeNode deseHelper(char[] data, int min, int max, int[] idx){
        if (idx[0]>=data.length || data[idx[0]]>max || data[idx[0]]return null;

        TreeNode root = new TreeNode(data[idx[0]]);
        idx[0]++;
        root.left = deseHelper(data, min, root.val, idx);
        root.right = deseHelper(data, root.val, max, idx);
        return root;
    }
}

AC代码3

特别聪明而且让人哭笑不得的代码：

public class Codec {
    TreeNode node;
    public String serialize(TreeNode root) {
        node=root;
        return "";
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        return node;
    }
}