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.
Example:
You may serialize the following tree:
1
/ \
2 3
/ \
4 5
as
"[1,2,3,null,null,4,5]"
Clarification:The above format is the same as. 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 & Analysis
Depth-first Search
Pre-Oder Traversal:
The preorder DFS traverse follows recursively the order of root -> left subtree -> right subtree.
用一个辅助的queue，存入data.split(",")之后的元素，这样在构建树时，就可以按照相同的顺序：root -> left subtree -> right subtree
Complexity Analysis
Time complexity : in both serialization and deserialization functions, we visit each node exactly once, thus the time complexity is O(N), whereNNis the number of nodes, _i.e._the size of tree.
Space complexity : in both serialization and deserialization functions, we keep the entire tree, either at the beginning or at the end, therefore, the space complexity is O(N).
10 ms, faster than 92.02%
/**
* 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 "";
}
StringBuilder sb = new StringBuilder();
traverse(root, sb);
// System.out.println(sb.toString());
return sb.toString();
}
private void traverse(TreeNode root, StringBuilder sb) {
if (root != null) {
sb.append(String.valueOf(root.val));
sb.append(",");
traverse(root.left, sb);
traverse(root.right, sb);
} else {
sb.append("#");
sb.append(",");
}
}
// Decodes your encoded data to tree.
public TreeNode deserialize(String data) {
if (data == null || data.isEmpty()) {
return null;
}
String[] values = data.split(",");
Queue<String> queue = new LinkedList<String>(Arrays.asList(values));
return buildTree(queue);
}
private TreeNode buildTree(Queue<String> queue) {
if (queue == null || queue.isEmpty()) {
return null;
}
String value = queue.poll();
if (value.equals("#")) {
return null;
}
TreeNode node = new TreeNode(Integer.parseInt(value));
node.left = buildTree(queue);
node.right = buildTree(queue);
return node;
}
}
// Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.deserialize(codec.serialize(root));
Concise Solution (serial part NOT RECOMMENDED)
1,2,#,#,3,4,#,#,5,#,#
优点在保持了末尾不会有多余",", 但是逻辑上比较缠绕，一会pass by reference，一会返回值。因此尽管代码比较简洁，不推荐在面试中使用。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Codec {
public String serialize(TreeNode root) {
return serial(new StringBuilder(), root).toString();
}
// Generate preorder string
private StringBuilder serial(StringBuilder str, TreeNode root) {
if (root == null) return str.append("#");
str.append(root.val).append(",");
serial(str, root.left).append(",");
serial(str, root.right);
return str;
}
public TreeNode deserialize(String data) {
return deserial(new LinkedList<>(Arrays.asList(data.split(","))));
}
// Use queue to simplify position move
private TreeNode deserial(Queue<String> q) {
String val = q.poll();
if ("#".equals(val)) return null;
TreeNode root = new TreeNode(Integer.valueOf(val));
root.left = deserial(q);
root.right = deserial(q);
return root;
}
}