0297. Serialize and Deserialize Binary Tree (H)

题目

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 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.

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.

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, 104].
• -1000 <= Node.val <= 1000

题意

实现二叉树的序列化和反序列化。

思路

按照前序遍历的顺序序列化二叉树，形式为结点的值用分隔符相连。反序列化同样参照前序遍历的顺序，递归处理即可。

代码实现

Java

public class Codec {
    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        if (root == null) {
            return "null";
        }
        String s = "" + root.val;
        s +="," + serialize(root.left);
        s += "," + serialize(root.right);
        return s;
    }
    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        String[] arr = data.split(",");
        Queue<TreeNode> q = new LinkedList<>();
        for (int i = 0; i < arr.length; i++) {
            q.offer(arr[i].equals("null") ? null : new TreeNode(Integer.parseInt(arr[i])));
        }
        return dfs(q);
    }
    private TreeNode dfs(Queue<TreeNode> q) {
        if (q.isEmpty()) {
            return null;
        }
        TreeNode cur = q.poll();
        if (cur == null) return null;
        cur.left = dfs(q);
        cur.right = dfs(q);
        return cur;
    }
}