297. Serialize and Deserialize Binary Tree[Hard]

297. Serialize and Deserialize Binary Tree

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.

Constraints:

• The number of nodes in the tree is in the range [0, 10^4].
• -1000 <= Node.val <= 1000

Example

Input: root = [1,2,3,null,null,4,5]
Output: [1,2,3,null,null,4,5]

思路

序列化的题，最主要的就是转成字符串后要用什么做标记，再转回去，这边val只能位数字，那就直接用空格来做标记了，如果是String的话，就要记录每个String的长度再加一个标志位

题解

    String result = "";
    Integer index = 0;

    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        dfsSerialize(root);
        return result;
    }

    public void dfsSerialize(TreeNode root) {
        if (root == null) {
	    // 为null就加上结束标记#再加一个空格
            result += "# ";
            return;
        }
	// 中左右 先序遍历 依次记录
        result += root.val + " ";
        dfsSerialize(root.left);
        dfsSerialize(root.right);
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
	// 利用标记位空格切分成数组，这样就可以直接取值，如果通过String取的话还要考虑数字长度是否为负数等等
        return dfsDeserialize(data.split(" "));
    }

    public TreeNode dfsDeserialize(String[] val) {
        if (val[index].equals("#")) {
	    // #为结束符，下标+1
            index++;
            return null;
        }
        TreeNode cur = new TreeNode();
	// 继续中左右 依次取出
        cur.val = Integer.parseInt(String.valueOf(val[index++]));
        cur.left = dfsDeserialize(val);
        cur.right = dfsDeserialize(val);
        return cur;
    }