[LeetCode] 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.

Example: 

You may serialize the following tree:

    1
   / 
  2   3
     / 
    4   5

as "[1,2,3,null,null,4,5]"

Clarification: The above 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.

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

二叉树的序列化和反序列化。题目即是题意，序列化是需要把二叉树convert成一个字符串；同时这个字符串也可以被convert回对应的二叉树。影子题449题。这道题有两种思路，一种是BFS层序遍历，一种是preorder前序遍历。

首先是BFS。序列化的部分是按照level order traversal的思路遍历树的节点。如果当前节点为NULL则存入"null"；若不是，则正常存入当前节点，再遍历当前节点的左孩子和右孩子。记得存入每个节点之后加一个空格，这样方便接下来的反序列。

反序列化是将一个字符串的input转换成树的结构。首先将字符串中第一个node拿到，当做根节点。之后还是根据BFS的思路遍历input，如果不为"null"则做成一个节点并加入queue；但是这里反序列化可以跳过null，不需要把null也转换成一个空的节点放入二叉树。

时间O(n)

空间O(n)

Java实现

public class Codec {

    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        if (root == null)
            return "";
        StringBuilder res = new StringBuilder();
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            if (cur == null) {
                res.append("null ");
                continue;
            }
            res.append(cur.val + " ");
            queue.offer(cur.left);
            queue.offer(cur.right);
        }
        return res.toString();
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if (data == "")
            return null;
        String[] str = data.split(" ");
        TreeNode root = new TreeNode(Integer.parseInt(str[0]));
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        for (int i = 1; i < str.length; i++) {
            TreeNode cur = queue.poll();
            if (!str[i].equals("null")) {
                cur.left = new TreeNode(Integer.parseInt(str[i]));
                queue.offer(cur.left);
            }
            i++;
            if (!str[i].equals("null")) {
                cur.right = new TreeNode(Integer.parseInt(str[i]));
                queue.offer(cur.right);
            }
        }
        return root;
    }
}

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

JavaScript实现

/**
 * Encodes a tree to a single string.
 *
 * @param {TreeNode} root
 * @return {string}
 */
var serialize = function (root) {
    let stack = [];
    let serialize = [];
    if (root == null) return [];
    stack.push(root);
    while (stack.length > 0) {
        let cur = stack.shift();
        serialize.push(cur ? cur.val : null);
        if (cur != null) {
            stack.push(cur.left);
            stack.push(cur.right);
        }
    }
    return serialize;
};

var deserialize = function (data) {
    if (data[0] == null) return null;
    let node = new TreeNode(data.shift());
    let queue = [node];
    while (queue.length > 0) {
        let node = queue.shift();
        left = data.shift();
        right = data.shift();
        node.left = (left == null) ? null : new TreeNode(left);
        node.right = (right == null) ? null : new TreeNode(right);
        if (node.left != null) {
            queue.push(node.left);
        }
        if (node.right != null) {
            queue.push(node.right);
        }
    }
    return node;
};

/**
 * Your functions will be called as such:
 * deserialize(serialize(root));
 */

再来是前序遍历的思路。既然是前序遍历，那么还是先处理当前节点，再处理他的左孩子和右孩子。序列化的部分，还是需要对空节点记录，我这里还是记录成"null" + 一个空格，然后调用helper函数递归处理左孩子和有孩子。反序列化的时候，除了一开始需要把data分开，同时需要把data分开的内容放入一个queue，这样每弹出一个元素，我们还是可以通过一个helper函数递归处理左孩子和右孩子节点。这个做法比之前BFS的做法稍稍快一点。

时间O(n)

空间O(n)

Java实现

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

    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        StringBuilder sb = new StringBuilder();
        helper(root, sb);
        return sb.toString();
    }

    private void helper(TreeNode root, StringBuilder sb) {
        if (root == null) {
            // sb.append("#").append(",");
            sb.append("null ");
        } else {
            sb.append(root.val).append(" ");
            helper(root.left, sb);
            helper(root.right, sb);
        }
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        // corner case
        if (data == null) {
            return null;
        }
        // normal case
        String[] strs = data.split(" ");
        Queue<String> queue = new LinkedList<>();
        Collections.addAll(queue, strs);
        return helper(queue);
    }

    private TreeNode helper(Queue<String> queue) {
        if (queue.isEmpty()) {
            return null;
        }
        String s = queue.poll();
        if (s.equals("null")) {
            return null;
        }
        TreeNode root = new TreeNode(Integer.parseInt(s));
        root.left = helper(queue);
        root.right = helper(queue);
        return root;
    }
}

// Your Codec object will be instantiated and called as such:
// Codec ser = new Codec();
// Codec deser = new Codec();
// TreeNode ans = deser.deserialize(ser.serialize(root));

相关题目

297. Serialize and Deserialize Binary Tree

428. Serialize and Deserialize N-ary Tree

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LeetCode 题目总结