Problem Description

The problem is about designing an algorithm to perform two complementary tasks: serializing a binary tree to a string and deserializing that string back to the original binary tree structure. Serialization implies converting the tree into a format that can be easily stored or transmitted, while deserialization is the process of reconstructing the tree from the serialized format.

There is flexibility in choosing the method of serialization and deserialization, as long as the serialized string can represent the structure of the tree accurately, and then the tree can be properly restored from this string representation.

It's important to note that the specific format LeetCode uses to represent trees is not mandatory. Developers are encouraged to be creative and design their own serialization/deserialization strategies.

Flowchart Walkthrough

To determine the most suitable algorithm for the LeetCode problem 297: Serialize and Deserialize Binary Tree, let's proceed step-by-step using the Flowchart. Here's how to analyze the problem:

Is it a graph?

• Yes: A binary tree is a specific type of graph, with nodes connected by edges where each node has at most two children.

Is it a tree?

• Yes: By definition, a binary tree is a tree structure.

Now that we've identified it as a tree:

• As per our flowchart, once it's confirmed as a tree, the further analysis on tree specifics would lead to the utilization of Depth-First Search (DFS) for traversal or manipulation. This aligns well with the typical patterns for serialization and deserialization in a tree struct, where DFS elegantly handles recursive traversal through branches (both left and right children).

Conclusion: The flowchart points directly to using DFS after confirming that it's a tree. Depth-First Search is perfect for serialization, as it allows capturing all details of a subtree before moving on, and for deserialization, to rebuild the tree by reading serialized data sequentially as it would have been generated by DFS. This makes it exceedingly suitable for the tasks of serializing and deserializing a binary tree.

Intuition

For the solution, we can use the well-known tree traversal methods. The chosen traversal method in this solution is preorder traversal, which processes the root before the nodes in the subtrees. This type of traversal assists in maintaining the hierarchy of the tree.

To serialize the tree:

1. We start with a preorder traversal, visiting nodes in a root-left-right order.
2. If a node is null, representing the absence of a child in the tree, we store a placeholder (in this case, a hash symbol "#").
3. Otherwise, we record the value of the node followed by a delimiter (a comma ",") to separate this node's value from the next.
4. After traversing the entire tree, we combine all the recorded values into a single string.

To deserialize the string back to a tree:

1. We first split the serialized string by the delimiter "," to get an array of values.
2. Then we follow the recorded order to reconstruct the tree. We use a recursive helper function that pops values from the array, starting from the first element.
3. If a value corresponds to the null placeholder, we return None to signify the absence of a node.
4. Otherwise, we create a new tree node with the current value and set its left and right children by making recursive calls to the helper function.

Since the preorder traversal approach is used, the position of each node is clearly defined by the order in which values appear in the serialized string. This order ensures that each call to the recursive helper function in the deserialization process constructs the correct tree structure.

Learn more about Tree, Depth-First Search, Breadth-First Search and Binary Tree patterns.

Solution Approach

The implementation uses the concept of preorder traversal for both serialization and deserialization. Let's break down the approach for each.

Serialization Process:

1. Preorder traversal recursion: When a node is visited, its value is recorded first, followed by its left and right children recursively.
2. Handling null nodes: Whenever the traversal encounters a None (e.g., a leaf has no left or right child), it appends a special symbol "#," to denote a null.
3. Data storage: The values (including null placeholders) are appended to a list, res, during each recursive call.
4. String conversion: After the traversal is complete, the list of values and placeholders are joined into a single string using ','.join(res), which is then returned as the serialization result.

The preorder traversal is particularly suitable for serialization because it encodes the structure of the tree starting from the root, following a top-down approach that corresponds to how a tree is naturally constructed.

Deserialization Process:

1. String splitting: The serialized string is split by the delimiter ',' to reconstruct the array of values and null placeholders.
2. Recursive reconstruction: The deserialization is also recursive, using a nested function inner() that reconstructs the tree by popping the first value from the array of values and using it to construct a TreeNode.
3. Null nodes handling: If the popped value is the null placeholder, inner() returns None to indicate the absence of a tree node.
4. Tree node creation: If the value is not the null placeholder, inner() creates a TreeNode with this value, and its left and right children are recursively set by subsequent calls to inner().

By consistently using preorder traversal, the deserialization process can follow the exact sequence of steps that were used to serialize the tree, ensuring that the original tree structure is accurately reconstructed.

Example parse tree for illustration:

Assume the serialization of a tree is "1,2,#,#,3,#,#,". The deserialization process would be:

• Start with 1: Create a node with value 1.
• Then move to 2: Create left child of 1 with value 2.
• Then "#,#": Indicates that 2 has no left or right child, so move back to the parent node 1.
• Now handle 3: Create right child of 1 with value 3.
• Finally, "#,#": Indicates that 3 has no left or right child, thus reconstruction is done.

By following this recursive approach, the tree's nodes are constructed in correct preorder, resulting in the tree being restored to its original structure.

Example Walkthrough

Let's consider a simple binary tree as an example to illustrate the serialization and deserialization process. Our binary tree structure will be as follows:

  1
 / \
2   3

Serialization:

We will serialize this tree using the serialization process defined above. Here's the step-by-step process:

1. Preorder traversal recursion: Start at the root node (1). Since it's not null, record its value. Then serialize its left subtree (2), followed by its right subtree (3).
2. Handling null nodes: As we serialize node 2, we find no left or right children, so we append "#," for each absent child. We do the same for node 3.
3. Data storage: Throughout the process, we keep appending the values or placeholders to our list res.
4. String conversion: After we traverse all the nodes, we join all the elements in res with commas to form the serialized string.

Following these steps, the serialization process for our tree will yield the string: "1,2,#,#,3,#,#,".

Deserialization:

Now, we'll take the serialized string "1,2,#,#,3,#,#," and convert it back into the original binary tree using the deserialization process:

1. String splitting: Split the string by ',' to get an array of values: ["1", "2", "#", "#", "3", "#", "#"].
2. Recursive reconstruction: Call the nested function inner() which will pop values from the array and construct the tree in preorder.
3. Null nodes handling: When inner() encounters a "#" (which is the null placeholder), it returns None. This indicates that the current node does not have a left or right child at that position in the tree.
4. Tree node creation: When inner() pops a numerical value, it creates a TreeNode with that value. It then sets the left child by making a recursive call to inner() (which will process the next value in the array), and the right child by another call to inner() after that.

Deserialization of our serialized string will create the root node 1, create a left child with value 2 (as the next value in the array is "2" and then encounter # placeholders, indicating that 2 has no children), move back to the root node, and then create a right child with value 3 (as the next value in the array is "3"). Following 3, we again encounter # placeholders indicating that 3 also does not have children. With no more elements in the array, the deserialization is complete.

As a result, the original binary tree is reconstructed:

  1
 / \
2   3

Throughout this example, you can see that the sequence of values and placeholders in the serialized string directly reflects the structure of the tree, allowing for precise reconstruction during deserialization.

Solution Implementation

Time and Space Complexity

serialize method

Time Complexity:

The time complexity of the serialize method is O(n), where n is the number of nodes in the tree. This is because the method performs a preorder traversal of the tree and processes each node exactly once.

Space Complexity:

The space complexity of the serialize method is O(n) as well. The reason for this is twofold:

1. The call stack during the recursion of the preorder function will at worst case be O(h) where h is the height of the tree. In the worst case of a skewed tree, h can be n.
2. The output list res will contain a total of 2n entries (including the # for None nodes), which is linearly proportional to the number of nodes in the tree, hence O(n).

deserialize method

Time Complexity:

The deserialize method also has a time complexity of O(n) because it iterates over each element of the input string once. The inner function is called for every node, meaning it runs n times.

Space Complexity:

The space complexity for the deserialize method is O(n) for the following reasons:

1. The split operation on the string creates a list of values vals which will be of size n (plus one for the trailing #).
2. The recursion stack depth of inner() will also be O(h). Just like in serialization, h can be n in the worst case, so space complexity due to recursion can be considered O(n).

Overall, when considering serialization and deserialization together, the time complexity is O(n) and the space complexity is O(n) for both operations.

Learn more about how to find time and space complexity quickly using problem constraints.