Understanding the Problem
We are given a binary tree, and our goal is to find all duplicate subtrees. A duplicate subtree is one where both the structure and node values are identical to another subtree in the tree. Each duplicate should be returned only once, represented by its root node.
To identify duplicates, we must compare every subtree with all others. But doing this naively would be inefficient. Instead, we use a serialization strategy to transform each subtree into a string format, allowing quick lookup and comparison using a hash map.
Step-by-Step Solution with Example
Step 1: Visualize the Example
Let’s take the input tree: [1,2,3,4,null,2,4,null,null,4]
This corresponds to the following structure:
1
/ 2 3
/ / 4 2 4
/
4
We can observe:
- Three nodes with value 4, all as leaf nodes.
- Two subtrees with structure
2 → 4
.
These are candidates for duplicate subtrees.
Step 2: Define Serialization Format
We perform a post-order traversal (left → right → node) and serialize each subtree into a string.
For example:
- Leaf node 4 becomes
"4,#,#"
- Subtree 2 → 4 becomes
"2,4,#,#,#"
This unique string helps us identify duplicates using a hash map that stores counts.
Step 3: Track and Detect Duplicates
We maintain a hash map where keys are serialization strings and values are their frequency.
As we serialize each subtree:
- If a serialization is seen once, we add the subtree root to the result.
- If it’s already seen more than once, we skip it to avoid duplicates in the result.
Step 4: Return the Result
From our example:
- Serialization
"4,#,#"
appears 3 times → add one root node of 4
- Serialization
"2,4,#,#,#"
appears 2 times → add one root node of 2
The final result contains the roots of these duplicate subtrees.
Edge Cases
Case 1: All Subtrees Are Unique
Input: [1,2,3]
Structure:
1
/ 2 3
Each node is unique. Serialized strings for each subtree do not match. Result:
[]
.
Case 2: Empty Tree
Input: []
There are no nodes in the tree. Hence, no subtrees exist. Result: []
.
Case 3: Large Tree with Many Duplicates
Ensure that your solution uses efficient traversal and hashing to handle such cases within time limits.
Finally
This problem is a great example of using serialization to simplify complex structure comparisons. It teaches you how to:
- Break down trees into reusable string representations
- Use hash maps to count and identify duplicates efficiently
- Handle edge cases like null roots and small trees gracefully
This technique is widely applicable in tree-related problems where comparison is required.
Comments
Loading comments...