题目描述

• 这道题涉及到不少 String、StringBuilder、Integer的转换、处理。
  

思路 && 代码

• 序列化：迭代进行一个层序遍历，逐个加入结果字符串中。
• 反序列化：根据序列化得到的结果字符串，同样是借助队列，进行层序遍历来构造出二叉树。
• 总的来说，是需要多敲几遍的题（String等类的用法）

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
public class Codec {// 知识点：String、StringBuilder、parseInt的处理// 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.add(root);// 层序遍历 BFS(迭代) while(!queue.isEmpty()) {TreeNode temp = queue.poll();if(temp != null) {// 当前值加入 res，子结点加入 queueres.append(temp.val).append(",");queue.add(temp.left);queue.add(temp.right);}// 空结点 情况else {res.append("null,");}}// 删除末尾的','res.delete(res.length() - 1, res.length());res.append("]");return res.toString();}// Decodes your encoded data to tree.public TreeNode deserialize(String data) {if(data.equals("[]")) {return null;}// 1. initString[] vals = data.substring(1, data.length() - 1).split(",");TreeNode root = new TreeNode(Integer.parseInt(vals[0]));Queue<TreeNode> queue = new LinkedList<>();queue.add(root);// 2. deserializefor(int i = 1; !queue.isEmpty(); i += 2) {TreeNode temp = queue.poll();// 左结点判断if(!vals[i].equals("null")) {temp.left = new TreeNode(Integer.parseInt(vals[i]));queue.add(temp.left);}// 右结点判断if(!vals[i + 1].equals("null")) {temp.right = new TreeNode(Integer.parseInt(vals[i + 1]));queue.add(temp.right);}}return root;}
}// Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.deserialize(codec.serialize(root));

二刷

• 序列化格式：逗号’,'作为分割符，“null” 作为空节点。按照层序遍历序列化
• 注意：序列化、反序列都需要队列辅助

public class Codec {public String serialize(TreeNode root) {if(root == null) {return "";}StringBuilder sb = new StringBuilder();LinkedList<TreeNode> queue = new LinkedList<>();queue.add(root);while(!queue.isEmpty()) {TreeNode temp = queue.poll();if(temp == null) {sb.append("null,");}else {sb.append(temp.val).append(',');queue.add(temp.left);queue.add(temp.right);}}return sb.toString();}public TreeNode deserialize(String data) {if(data.length() == 0) {return null;}String[] vals = data.split(",");TreeNode root = new TreeNode(Integer.parseInt(vals[0]));LinkedList<TreeNode> queue = new LinkedList<>(); // 还是需要辅助队列queue.add(root);for(int i = 1; i < vals.length; i += 2) {TreeNode temp = queue.poll();if(!vals[i].equals("null")) {temp.left = new TreeNode(Integer.parseInt(vals[i]));queue.add(temp.left);}if(!vals[i + 1].equals("null")) {temp.right = new TreeNode(Integer.parseInt(vals[i + 1]));queue.add(temp.right);}}return root;}
}