# Serialize and Deserialize Binary Tree

`Tree`, `Design`

**Hard**

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[how LeetCode serializes a binary tree](https://leetcode.com/faq/#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 & Analysis

### Depth-first Search

**Pre-Oder Traversal**:

The preorder DFS traverse follows recursively the order of **root -> left subtree -> right subtree**.

```
[1,2,3,null,null,4,5]

---serialize--->

1,2,#,#,3,4,#,#,5,#,#,
```

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%

```java
/**
 * 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)

via [@cdai](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/discuss/74253/Easy-to-understand-Java-Solution/77363) -- 以下：代码比较简洁，但是serial的部分不是很readable的地方在于，何时加delimiter `","`

```
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**。

```java
/**
 * 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;
    }
}
```

### Breadth-first Search - Queue

Serialize时类似 **pre-order traversal**. 这里为了节省空间，删去了末尾的"#"，虽然deserialized的时候并无影响。

Deserialize的时候要从左边节点开始填充，也就是说只有i % 2 == 1时才从queue中poll()，如果不为null，再生成新节点放入queue中。

```java
/**
 * 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();
        // sb.append('[');

        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            if (node == null) {
                sb.append('#');
            } else {
                sb.append(node.val);
                queue.offer(node.left);
                queue.offer(node.right);
            }
            sb.append(',');
        }

        // (optional, won't affect deserialization) remove tailing sharp and comma
        for (int i = sb.length() - 1; i >= 0; i--) {
            if (sb.charAt(i) == '#' || sb.charAt(i) == ',') {
                sb.deleteCharAt(i);
            } else {
                break;
            }
        }

        // sb.append(']');
        return sb.toString();
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if (data == null || data.isEmpty()) {
            return null;
        }
        String[] values = data.split(",");
        TreeNode root = new TreeNode(Integer.parseInt(values[0]));
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        q.offer(root);

        int i = 1;
        boolean isLeft = true;
        TreeNode node = null;
        while (i < values.length) {
            if (isLeft) {
                node = q.poll();
                if (values[i].equals("#")) {
                    node.left = null;
                } else {
                    node.left = new TreeNode(Integer.parseInt(values[i]));
                    q.offer(node.left);
                }
            } else {
                if (values[i].equals("#")) {
                    node.right = null;
                } else {
                    node.right = new TreeNode(Integer.parseInt(values[i]));
                    q.offer(node.right);
                }
            }
            isLeft = !isLeft;
            i++;
        }
        return root;
    }
}

// Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.deserialize(codec.serialize(root));
```

## Reference

<https://leetcode.com/problems/serialize-and-deserialize-binary-tree/solution/>

<https://leetcode.com/problems/serialize-and-deserialize-binary-tree/discuss/74253/Easy-to-understand-Java-Solution>

<https://www.jiuzhang.com/solutions/binary-tree-serialization/#tag-other>

<https://leetcode.com/problems/serialize-and-deserialize-binary-tree/discuss/74260/Recursive-DFS-Iterative-DFS-and-BFS>