Hard 297题 Serialize and Deserialize Binary Tree
Question:
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.
Solution:
/**
* 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) {
if (root == null) return "";
Queue q=new LinkedList();
StringBuilder res = new StringBuilder();
q.offer(root);
while(!q.isEmpty())
{
TreeNode tmp=q.poll();
if(tmp!=null)
res=res.append(tmp.val).append(",");
else
{
res=res.append("null,");
continue;
}
q.offer(tmp.left);
q.offer(tmp.right);
}
return res.toString().substring(0,res.length()-1);
}
// Decodes your encoded data to tree.
public TreeNode deserialize(String data) {
if (data == null||data.length()==0)
return null;
String[] splitInfo = data.split(",");
TreeNode res=new TreeNode(Integer.parseInt(splitInfo[0]));
Queue q=new LinkedList();
q.offer(res);
for(int i=1;i<=splitInfo.length-1;i++)
{
TreeNode tmp_parent=q.poll();
String tmp_1=splitInfo[i];
if(!tmp_1.equals("null")){
tmp_parent.left=new TreeNode(Integer.parseInt(splitInfo[i]));
q.offer(tmp_parent.left);
}
String tmp_2=splitInfo[++i];
if(!tmp_2.equals("null")){
tmp_parent.right=new TreeNode(Integer.parseInt(splitInfo[i]));
q.offer(tmp_parent.right);
}
}
return res;
}
}
// Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.deserialize(codec.serialize(root)); DFS11/17
/**
* 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();
DFSserialize(root,sb);
return sb.toString().substring(0, sb.toString().length()-1);
}
public void DFSserialize(TreeNode node,StringBuilder sb){
if(node==null){
sb.append("null,");
return;
}
sb.append(node.val).append(",");
DFSserialize(node.left, sb);
DFSserialize(node.right,sb);
}
// Decodes your encoded data to tree.
public TreeNode deserialize(String data) {
if(data.length()==0||data==null)
return null;
String[] split=data.split(",");
Queue q=new LinkedList();
for(int i=0;i<=split.length-1;i++){
q.offer(split[i]);
}
return DFSdeserialize(q);
}
public TreeNode DFSdeserialize(Queue q){
String val = q.poll();
if (val.equals("null"))
return null;
else {
TreeNode node=new TreeNode(Integer.parseInt(val));
node.left=DFSdeserialize(q);
node.right=DFSdeserialize(q);
return node;
}
}
}
// Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.deserialize(codec.serialize(root));