LC.P449[序列化和反序列化二叉搜索树]

方法一：BFS

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
/**
 * 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 builder = new StringBuilder();
        Deque<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            if (cur == null) {
                builder.append("#").append(",");
            } else {
                builder.append(cur.val).append(",");
                queue.offer(cur.left);
                queue.offer(cur.right);
            }
        }
        return builder.toString();
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if (data.isEmpty()) return null;
        Deque<String> nodes = new LinkedList<>(List.of(data.split(",")));
        TreeNode root = new TreeNode(Integer.parseInt(nodes.poll()));
        Deque<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            String left = nodes.poll();
            String right = nodes.poll();
            if (!left.equals("#")) {
                cur.left = new TreeNode(Integer.parseInt(left));
                queue.offer(cur.left);
            }
            if (!right.equals("#")) {
                cur.right = new TreeNode(Integer.parseInt(right));
                queue.offer(cur.right);
            }
        }
        return root;
    }
}

// Your Codec object will be instantiated and called as such:
// Codec ser = new Codec();
// Codec deser = new Codec();
// String tree = ser.serialize(root);
// TreeNode ans = deser.deserialize(tree);
// return ans;
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$

方法二：先序遍历

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
/**
 * 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 "#,";
        String left = serialize(root.left);
        String right = serialize(root.right);
        return root.val + "," + left + right;
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if (data.isEmpty()) return null;
        Deque<String> queue = new LinkedList<>(List.of(data.split(",")));
        return buildTree(queue);
    }

    private TreeNode buildTree(Deque<String> queue) {
        String cur = queue.poll();
        if (cur.equals("#")) return null;
        TreeNode root = new TreeNode(Integer.parseInt(cur));
        root.left = buildTree(queue);
        root.right = buildTree(queue);
        return root;
    }
}

// Your Codec object will be instantiated and called as such:
// Codec ser = new Codec();
// Codec deser = new Codec();
// String tree = ser.serialize(root);
// TreeNode ans = deser.deserialize(tree);
// return ans;
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$