题目
二叉树的序列化和反序列化
设计一个算法，并编写代码来序列化和反序列化二叉树。将树写入一个文件被称为“序列化”，读取文件后重建同样的二叉树被称为“反序列化”。
如何反序列化或序列化二叉树是没有限制的，你只需要确保可以将二叉树序列化为一个字符串，并且可以将字符串反序列化为原来的树结构。
样例
给出一个测试数据样例， 二叉树{3,9,20,#,#,15,7}，表示如下的树结构：
3
/ \
9 20
/ \
15 7
我们的数据是进行BFS遍历得到的。当你测试结果wrong answer时，你可以作为输入调试你的代码。
你可以采用其他的方法进行序列化和反序列化。
解题
参考九章程序
看看注释就理解了，但是我表示自己想不出来Java
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
class Solution {
/**
* This method will be invoked first, you should design your own algorithm
* to serialize a binary tree which denote by a root node to a string which
* can be easily deserialized by your own "deserialize" method later.
*/
public String serialize(TreeNode root) {
// write your code here
if( root == null)
return "{}";
ArrayList<TreeNode> queue = new ArrayList<TreeNode>();
queue.add(root);
// 将二叉树的个节点按照从上到下、从左到有的存储在queue中
for(int i=0;i<queue.size();i++){
TreeNode q = queue.get(i);
if(q== null)
continue;
queue.add(q.left);
queue.add(q.right);
}
// 去除叶子节点的左右孩子，这个孩子是空值
while(queue.get(queue.size() - 1) == null){
queue.remove(queue.size() - 1);
}
// 遍历queue把转换成字符串
StringBuilder sb = new StringBuilder();
sb.append("{");
sb.append(queue.get(0).val);
for(int i=1;i<queue.size(); i++){
TreeNode q = queue.get(i);
if(q!= null){
sb.append(",");
sb.append(q.val);
}else{
sb.append(",#");
}
}
sb.append("}");
return sb.toString();
} /**
* This method will be invoked second, the argument data is what exactly
* you serialized at method "serialize", that means the data is not given by
* system, it's given by your own serialize method. So the format of data is
* designed by yourself, and deserialize it here as you serialize it in
* "serialize" method.
*/
public TreeNode deserialize(String data) {
// write your code here
if(data == "{}")
return null;
// 以逗号分割
String[] vals = data.substring(1,data.length()-1).split(",");
ArrayList<TreeNode> queue = new ArrayList<TreeNode>();
// 根节点
TreeNode root = new TreeNode(Integer.parseInt(vals[0]));
queue.add(root);
int index = 0;
boolean isLeftChild = true;
for (int i = 1; i < vals.length; i++) {
if (!vals[i].equals("#")) {
TreeNode node = new TreeNode(Integer.parseInt(vals[i]));
if (isLeftChild) {
queue.get(index).left = node;
} else {
queue.get(index).right = node;
}
queue.add(node);
}
if (!isLeftChild) {
index++;
}
isLeftChild = !isLeftChild;
}
return root;
}
}