题目要求
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.
For example, you may serialize the following tree
1
/ \
2 3
/ \
4 5
as "[1,2,3,null,null,4,5]", just the same as how LeetCode OJ 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.
设计一个方法将一个二叉树序列化并反序列化。序列化是指将对象转化成一个字符串,该字符串可以在网络上传输,并且到达目的地后可以通过反序列化恢复成原来的对象。
在Leetcode中,树的序列化是通过广度优先遍历实现的,就如题目中所介绍的那样。
这里我们使用中序遍历来解决这个问题。
思路和代码
中序遍历是指递归的先遍历当前节点,然后按同样的方式递归的遍历左子树,再递归的遍历右子树。因此题目中的例子的中序遍历结果为[1,2,3,4,5]
。但是,标准的中序遍历并不能使我们将该结果反构造成一棵树,我们丢失了父节点和子节点之间的关系。因此我们需要适当的插入空值来保存父节点和子节点的关系。
因此,我们将访问到空值也记录下来,结果如下[1,2, , ,3,4, , ,5, , ]
。
这里我们也可以明显的看出来,中序遍历需要保存的空值远远多于广度优先遍历。
那么我们如何将这个字符串反序列化呢?
其实思路还是一样的,将其根据分隔符分割后传入队列,不断从队列头读取数据,先构建当前节点,然后依次构建左子树和右子树,如果遇到空值,则返回递归。
代码如下:
private static final String NULL = "N";
private static final String SPLITOR = ",";
// Encodes a tree to a single string.
public String serialize(TreeNode root) {
StringBuilder result = new StringBuilder();
inOrder(root, result);
return result.toString();
}
private void inOrder(TreeNode root, StringBuilder result){
if(root==null){
result.append(NULL).append(SPLITOR);
}else{
result.append(root.val).append(SPLITOR);
inOrder(root.left, result);
inOrder(root.right, result);
}
}
// Decodes your encoded data to tree.
public TreeNode deserialize(String data) {
Deque<String> nodes = new LinkedList<String>(Arrays.asList(data.split(SPLITOR)));
return buildTree(nodes);
}
private TreeNode buildTree(Deque<String> nodes) {
String val = nodes.remove();
if (val.equals(NULL)) return null;
else {
TreeNode node = new TreeNode(Integer.valueOf(val));
node.left = buildTree(nodes);
node.right = buildTree(nodes);
return node;
}
}
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