Find Duplicate Subtrees
Problem Description
Given the root of a binary tree, return all duplicate subtrees.
For each kind of duplicate subtrees, you only need to return the root node of any one of them.
Two trees are duplicate if they have the same structure with the same node values.
Example 1:
Output: [[2,4],[4]]
Example 2:
Output: [[1]]
Example 3:
Output: [[2,3],[3]]
Constraints:
- The number of the nodes in the tree will be in the range
[1, 5000] -200 <= Node.val <= 200
Solve it on LeetCode
Approaches
1. Brute Force
Intuition:
The brute force approach involves comparing each subtree with every other subtree. We can implement this by conducting a traversal of the tree and checking each subtree for duplicates. However, this might replicate work since the same structure could be recomputed multiple times.
Steps:
- Traverse each node in the tree.
- Extract the subtree starting from that node.
- Compare the extracted subtree with all other subtrees in the tree to check for duplicates.
- Store and return the duplicate subtrees.
This approach is inefficient because of the repeated subtree comparisons.
Complexity Analysis:
- Time Complexity:
O(n^2*m), wherenis the number of nodes andmis the average number of nodes in each subtree. - Space Complexity:
O(n*m), for storing subtrees and comparisons.
2. Serialization with HashMap
Intuition:
Instead of comparing subtrees directly, we can serialize each subtree into a string and use a hashmap to count the frequency of each serialized subtree. This approach allows us to efficiently check if we've seen the same subtree structure before.
Steps:
- Serialize each subtree rooted at each node into a string.
- Use a HashMap to keep track of the frequency of each serialized subtree.
- If the frequency of a serialized subtree is greater than
1, it indicates a duplicate. - Keep track of the first instance of such duplicates.
Code:
- Time Complexity:
O(n), because we traverse each node once and serialization isO(1)for each node. - Space Complexity:
O(n), for storing serialized subtrees in the hashmap.
3. Optimized Serialization with Unique ID
Intuition:
In the optimized approach, instead of using strings to serialize each subtree, we assign a unique ID to each distinct subtree structure. This reduces the space complexity as we avoid storing large strings.
Steps:
- Use a map to assign a unique ID to each distinct subtree structure.
- Maintain a second map to count the occurrence of these unique IDs.
- If a subtree's unique ID occurs more than once, it indicates a duplicate subtree.
Code:
- Time Complexity:
O(n), traversing each node once and using constant time operations for map manipulations. - Space Complexity:
O(n), since we maintain maps to record unique IDs for the subtrees.