# LeetCode Serialize and Deserialize Binary Tree (Java)

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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

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.

## Analysis

Based on the hint, we can use the same method of how LeetCode OJ serializes a binary tree to serialize a binary tree based on the idea of level order tree traversal.

For example, the above tree will result in this string `S = "1,2,3,null,null,4,5"`

In the deserialization stage, we know the first element (`1`) is the `root` of the tree. The second element (`2`) is the root’s left child, denoted as `rootLeft`, the third element (`3`) is the root’s right child, denoted as `rootRight`. Then the following two elements are the left and right child of the node `rootLeft`

If we process the elements in `S` one by one. In the first step, we can easily create the root Node (e.g, `root` = new TreeNode(1)).

Then we scan the remaining elements in `S`, and use the first two elements (e.g. `2 and 3`) to build the left and right child of `root`

Then we come to the `4th` and `5th` element (both are null). We know they are the children of `rootLeft`, but how can we get the reference of `rootLeft` when we encounter the `4th and 5th` element in `S`?

We can use a Queue to store `rootLeft` as soon as it is created. So when we encounter the `4th` and `5th` element in `S`, we just poll the head of the queue, which is `rootLeft`

See the following code for how we use queue in the deserialization stage.