Problem Description

In this problem, we are given a binary tree and we have to identify all the subtrees that appear more than once within that tree. A subtree consists of a node and all of its descendants. Subtrees are considered duplicates if they have identical structure and node values. That is to say, if you took two duplicate subtrees and compared them, they would be indistinguishable in terms of both their shape and the values held in each node. The problem requires returning a list where each element is the root node of one of the duplicate subtrees. If there are multiple duplicates of the same subtree, we only need to include one instance of it in our answer.

Flowchart Walkthrough

Let's determine the appropriate algorithm for solving Leetcode 652. Find Duplicate Subtrees, using the Flowchart. Below is a detailed explanation through the flowchart:

1. Is it a graph?

   • Yes: In this problem, trees are graphs without cycles, and subtrees can be considered as smaller graphs.

2. Is it a tree?

   • Yes: The main data structure discussed in the problem is a binary tree.

3. DFS

   • The decision leads directly to applying Depth-First Search.

Conclusion: The flowchart dictates the use of DFS for solving the problem where we need to traverse a tree to find duplicate subtrees. DFS is well-suited as it allows exploring each node and its children thoroughly which is necessary to serialize or encode subtrees for comparison.

Intuition

The core idea of finding duplicate subtrees is to use a traversal algorithm to serialize or represent each subtree in a unique way so that we can compare them directly. Serialization means converting the structure of a subtree into a string (or any other form of a data structure that can be used for comparison purposes) which, if two serialized strings are the same, ensures that the subtrees they represent are identical.

A common method is to use a post-order depth-first search (DFS) to serialize the tree. Post-order traversal means we visit a node's left child, then its right child, and finally the node itself. We define a helper function dfs to perform the traversal and serialize the subtrees as it recursively processes the tree. For each node, we create a string that includes the node's value and the serialization of its left and right children. This way, identical subtrees yield the same serialized string.

We store the frequency of each serialized subtree in a counter (a hashmap), and if we come across the same serialization again, it means we have found a duplicate subtree. We keep a list called ans where we add the root of a subtree when we encounter its serialization the second time (when the count becomes 2). We continue this process until every subtree has been serialized and processed, ending up with a list of roots for all duplicate subtrees.

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

Solution Approach

The proposed solution involves implementing a post-order depth-first search (DFS) to traverse the tree, serialize each subtree, and use a counter to detect duplicates. The steps in the implementation and the data structures involved are described below:

1. Define a helper function dfs(root) that will:

   • Take the current root of a subtree as an input.
   • Base case: If the current root is None, return a special placeholder ('#') to indicate a null node.
   • Recur for the left child and right child, serializing both subtrees.
   • Construct the serialized version of the current subtree, denoted by v, using the value of the current node and the representations of the left and right subtrees in a comma-separated format (e.g., val,left_repr,right_repr).

2. Use a Counter, which is a special dictionary from Python's collections module that keeps track of the frequency of each serialized subtree representation.

3. In the main findDuplicateSubtrees method, define an empty list ans to store the roots of detected duplicate subtrees.

4. Call the helper function dfs(root) to start the process from the root of the entire tree.

5. Within the dfs function, after getting the serialized representation of the current subtree:

   • Increment the count for this serialized form in the counter.
   • If the count is exactly 2, it means we've found a duplicate for the first time, so we append the current root to ans (we only add the subtrees when their count hits 2 to ensure we don't add duplicates of duplicates).

6. Once the DFS and serialization process completes, return the list ans which now contains the roots of all duplicate subtrees.

The choice of using a post-order DFS is strategic for a couple of reasons:

• When you process a node, its left and right subtrees have already been serialized and compared, so the comparison of the structure (shape) is naturally included.
• The construction of strings for serialization allows an accurate and efficiently comparable representation of each subtree, making it easy to use a counter to detect duplicates.

By using serialization and a hash-based counter, this approach ensures that the resulting list contains each duplicate subtree exactly once, following the problem's constraints.

Example Walkthrough

Let's use a small binary tree to illustrate the solution approach:

Consider the following binary tree:

    1
   / \
  2   3
 /   / \
4   2   4
   /
  4

In this tree, the leftmost subtree 2 -> 4 appears again as the left subtree of node 3. We'll walk through the solution steps to find this duplicate subtree.

1. We define a helper function dfs which takes a node of the binary tree and serializes the subtree rooted at that node.

2. Starting with the root, dfs performs post-order traversal. This means it will first process all nodes in the left subtree, then the right subtree, and finally, the current node.

3. At each node, the dfs function:

   • Returns the string '#' if the node is None.
   • Calls itself recursively to get the serialized string of the left subtree.
   • Calls itself recursively to get the serialized string of the right subtree.
   • Combines the value of the current node and the serialized strings from step 3 in a comma-separated format. For the root node '1', assuming its left serialization is '2,#,4' and right is '3,2,#,4,#,4', the serialized string would be '1,2,#,4,3,2,#,4,#,4'.

4. We use a Counter to track the occurrence of each serialized subtree. Initially it is empty.

5. Whenever we serialize a subtree in our dfs function:

   • We add the serialization to the Counter.
   • If we come across a serialized subtree that already has a count of 1, which means this is the second time we have found this subtree, we add its root node to the answer list ans.

Following the traversal, we end up with the following serializations after dfs function is completed:

• 4 (for the leaves with value 4)
• 2,#,4 (for the subtrees rooted at the nodes with value 2)
• 3,2,#,4,#,4 (for the subtree rooted at the node with value 3)
• 1,2,#,4,3,2,#,4,#,4 (for the whole tree)

The 'Counter' at the end of traversal would have the serialized strings with their frequencies:

• '#': 3 (placeholder for null, for each node without one child or more)
• 4: 3 (for each leaf node)
• 2,#,4: 2 (for the subtrees that appear twice)
• The rest will have a count of 1 because they don't have duplicates.

Since 2,#,4 has a frequency of 2, we only add the root node of its first instance to the ans list, which contains the node with the value '2' that is the left child of the root node '1'.

6. At the end of traversal, the ans list contains the root node of the duplicate subtree found, which is just the node with value '2' in our example.

After we have finished traversing the entire tree, the solution would be the node with value '2' that appeared as the left child of root '1', because the structure 2 -> 4 is duplicated in the tree.

Solution Implementation

Time and Space Complexity

Time Complexity:

The main operation in the provided code is a Depth-First Search (DFS) through the binary tree which involves visiting every node exactly once and constructing a serialization string for each node by recursively obtaining the serialization of its left and right children. The time complexity for visiting all nodes in the tree is O(N), where N is the number of nodes in the tree. However, constructing the serialization string for a node involves concatenation operations, which in Python have a time complexity that depends on the length of the strings being concatenated. In the worst case, these strings can get as large as O(N) when the tree degenerates into a linked list, with N being the number of nodes. This concatenation happens for every node, making the time complexity O(N^2) in the worst case.

Additionally, the serialization string for each node is stored in a hash table (counter), and each insertion operation can be considered O(1) on average, though in the worst case, due to hash collisions, it could be O(N). However, this is rare and typically the average case is expected.

Therefore, the average time complexity of the DFS is O(N^2) due to the serialization concatenation, but it could potentially degrade to O(N^3) if every hash insert operation takes O(N) time, which is a very pessimistic assumption and unlikely in practice.

Space Complexity:

The space complexity of the algorithm comes from three main sources:

1. The recursion stack used by DFS, which goes as deep as the height of the tree. In the worst case (degenerate tree), this height is O(N), contributing O(N) to the space complexity.

2. The counter hash table which stores the serialization strings for nodes. In the worst case, every node has a unique serialization, so the space used by the counter can grow up to O(N^2) since every serialization can be of length O(N).

3. The list of answers (ans), which holds a reference to each node that has a duplicate subtree. Since each node could potentially be a part of this list just once, this contributes O(N) to space complexity.

Considering these parts, the overall space complexity of the solution is O(N^2) due to the storage required for the serialization strings in the hash table.

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