297. Serialize and Deserialize Binary Tree
Problem
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.
Clarification: The input/output 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.
Example 1:
Input: root = [1,2,3,null,null,4,5]
Output: [1,2,3,null,null,4,5]
Example 2:
Input: root = []
Output: []
Example 3:
Input: root = [1]
Output: [1]
Example 4:
Input: root = [1,2]
Output: [1,2]
Code
449, 652
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 node = queue.poll();
if (node == null) {
res.append("null ");
continue;
}
res.append(node.val + " ");
queue.offer(node.left);
queue.offer(node.right);
}
return res.toString();
}
// Decodes your encoded data to tree.
public TreeNode deserialize(String data) {
if (data == null || data.length() == 0) 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 node = queue.poll();
if (!str[i].equals("null")) {
node.left = new TreeNode(Integer.parseInt(str[i]));
queue.offer(node.left);
}
i++;
if (!str[i].equals("null")) {
node.right = new TreeNode(Integer.parseInt(str[i]));
queue.offer(node.right);
}
}
return root;
}
}