Serialize and Deserialize Binary Tree
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.
Tips:
just use comma and null, BFS, O(n)
Code:
public String serialize(TreeNode root) {
if (root == null) {
return "null";
}
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
StringBuilder result = new StringBuilder();
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode current = queue.poll();
if (current == null) {
result.append("null,");
continue;
}
result.append(current.val).append(",");
queue.offer(current.left);
queue.offer(current.right);
}
}
System.out.println(result);
return result.toString();
}
// Decodes your encoded data to tree.
public TreeNode deserialize(String data) {
String[] nodes = data.split(",");
int index = 0;
if (nodes[index].equals("null")) {
return null;
}
TreeNode root = new TreeNode(Integer.parseInt(nodes[index++]));
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode current = queue.poll();
if (!nodes[index].equals("null")) {
current.left = new TreeNode(Integer.parseInt(nodes[index]));
queue.offer(current.left);
}
index++;
if (!nodes[index].equals("null")) {
current.right = new TreeNode(Integer.parseInt(nodes[index]));
queue.offer(current.right);
}
index++;
}
}
return root;
}