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Find Duplicate Subtrees

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TreesRecursionStrings

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:

Input: root = [1,2,3,4,null,2,4,null,null,4]
Output: [[2,4],[4]]

Example 2:

Input: root = [2,1,1]
Output: [[1]]

Example 3:

Input: root = [2,2,2,3,null,3,null]
Output: [[2,3],[3]]

Constraints:

The number of the nodes in the tree will be in the range [1, 5000]
-200 <= Node.val <= 200

Try coding it in LeetCode

Find Duplicate Subtrees

Medium

Asked by:

12 views

Topics:

TreesRecursionStrings

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:

Input: root = [1,2,3,4,null,2,4,null,null,4]
Output: [[2,4],[4]]

Example 2:

Input: root = [2,1,1]
Output: [[1]]

Example 3:

Input: root = [2,2,2,3,null,3,null]
Output: [[2,3],[3]]

Constraints:

The number of the nodes in the tree will be in the range [1, 5000]
-200 <= Node.val <= 200

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the node data within the tree?
Can the tree be empty or contain only a single node?
How should subtree equality be determined? Should the node values and the tree structure be identical for two subtrees to be considered duplicates?
If multiple duplicate subtrees exist, should I return all of them, or just one representative subtree for each unique duplicate?
What is the expected output format? Specifically, should I return the root nodes of the duplicate subtrees, or some other representation?

Brute Force Solution

Approach

We want to find all the trees within a larger tree that look exactly the same. The brute force method is like taking every possible subtree and comparing it to every other subtree to see if they match.

Here's how the algorithm would work step-by-step:

First, consider every single part of the tree as a potential subtree.
For each of these subtrees, create a description that captures its structure and the information it holds.
Now, take the first subtree and compare its description to the description of every other subtree.
If two subtrees have the exact same description, then we have found a duplicate.
Repeat this comparison process for every subtree against all other subtrees.
Keep a record of each unique duplicate subtree that you find to avoid reporting the same duplicate multiple times.

Code Implementation

def find_duplicate_subtrees_brute_force(root):
    duplicate_subtrees = []
    all_subtrees = []

    def traverse(node):
        if not node:
            return

        all_subtrees.append(node)
        traverse(node.left)
        traverse(node.right)

    traverse(root)

    def subtree_description(node):
        if not node:
            return "#"
        return str(node.val) + "," + subtree_description(node.left) + "," + subtree_description(node.right)

    for i in range(len(all_subtrees)):
        for j in range(i + 1, len(all_subtrees)):
            # Compare all possible subtrees.
            if subtree_description(all_subtrees[i]) == subtree_description(all_subtrees[j]):

                is_duplicate = False
                # Check if we've already recorded the duplicate.
                for previously_found in duplicate_subtrees:
                    if subtree_description(previously_found) == subtree_description(all_subtrees[i]):
                        is_duplicate = True
                        break

                if not is_duplicate:
                    # Record the duplicate if it's a new one.
                    duplicate_subtrees.append(all_subtrees[i])

    return duplicate_subtrees

Big(O) Analysis

Time Complexity

O(n^2 * m) – The algorithm considers each of the n nodes in the tree as the root of a potential subtree. For each of these n subtrees, it generates a description of the subtree, which takes O(m) time where m is the size of the subtree. Then, each subtree is compared to every other subtree which requires comparing n subtrees with each other. The comparison process, checking the description equality, has a cost proportional to size of subtree m. Therefore the overall complexity is approximately O(n * n * m), thus O(n^2 * m).

Space Complexity

O(N^2) – The algorithm considers every node as a potential subtree, leading to N subtrees where N is the number of nodes in the tree. For each subtree, a description (likely a string) needs to be stored. In the worst-case where each subtree is unique, storing these descriptions requires O(N) space each, resulting in O(N^2) space for all subtree descriptions. Additionally, a record of unique duplicate subtrees needs to be kept, which in the worst case (many duplicates) could require storing descriptions of at most N subtrees for the final result, which is still bounded by O(N^2).

Optimal Solution

Approach

The efficient way to find duplicate subtrees is to represent each subtree as a unique string. We can then compare these strings to identify duplicates without traversing the entire tree multiple times. This avoids redundant checks and makes the process much faster.

Here's how the algorithm would work step-by-step:

First, create a way to turn each subtree into a string representation. This string should capture the structure and data of the entire subtree.
For any empty spot on the tree (like a missing branch), use a special marker in the string, so they are all consistently handled.
Now, go through the entire tree, turning each subtree into its string form. Store each unique string we encounter.
If we find a string that's already stored, that means we've found a duplicate subtree. Keep track of the root node of the duplicate subtree.
In the end, report all the root nodes of the duplicate subtrees we've found.

Code Implementation

def find_duplicate_subtrees(root):
    subtree_representation_map = {}
    duplicate_subtrees = []

    def get_subtree_string(node):
        if not node:
            return "#"

        # Construct a unique string for the subtree.
        subtree_string = str(node.val) + "," + get_subtree_string(node.left) + "," + get_subtree_string(node.right)

        # Check if the subtree string already exists.
        if subtree_string in subtree_representation_map:
            subtree_representation_map[subtree_string] += 1

            # Add the node to duplicates only once.
            if subtree_representation_map[subtree_string] == 2:

                # Mark this node as the root of a duplicate.
                duplicate_subtrees.append(node)
        else:

            # Store the string if it's the first time.
            subtree_representation_map[subtree_string] = 1

        return subtree_string

    get_subtree_string(root)

    # Return all duplicate subtree root nodes.
    return duplicate_subtrees

Big(O) Analysis

Time Complexity

O(n^2) – The algorithm traverses each of the n nodes in the tree to serialize the subtree rooted at that node. Serializing a subtree potentially involves visiting all its nodes, but in the worst case, this could approach n nodes for each of the n subtrees. Further, checking for duplicate serialized subtrees requires comparing the current subtree's string representation against previously seen string representations, which, in the worst case, requires comparing against n subtree string representations. Hence, the time complexity is driven by nested operations that serialize and compare all subtrees, leading to approximately n * n operations, which simplifies to O(n^2).

Space Complexity

O(N^2) – The algorithm stores string representations of subtrees in a hash map to detect duplicates. In the worst case, where all subtrees are unique, the hash map stores N strings, where N is the number of nodes in the tree. Each string, representing a subtree, can have a length proportional to the number of nodes in that subtree, which in the worst case can be O(N). Therefore, storing N strings of length O(N) results in O(N * N) = O(N^2) space. The recursion stack also contributes to the space complexity, but its contribution of O(H) where H is the height of the tree is dominated by O(N^2) in the worst case.

Edge Cases

Case	How to Handle
Null or empty tree as input	Return an empty list immediately since there are no subtrees to compare.
A single-node tree	Return an empty list as a single node tree cannot have duplicate subtrees.
A tree where all nodes have the same value	The entire tree and all subtrees will be duplicates; ensure only the root of each distinct subtree structure is added to the result.
A tree where all nodes have distinct values	Return an empty list because no duplicate subtrees will exist.
Very large tree (potential stack overflow with recursion)	Consider using an iterative approach to traverse the tree instead of recursion, or increase stack size if using recursion.
Deeply skewed tree (e.g., a linked list)	This edge case tests the performance of the algorithm with imbalanced trees, potentially leading to worst-case time complexity; optimize the subtree comparison for such cases.
Trees with identical structure but different node values	The serialization/hashing method needs to incorporate node values to distinguish these trees.
Tree contains duplicate subtrees that are not direct children	The algorithm should identify all duplicate subtrees regardless of their level in the tree.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the node data within the tree?
Can the tree be empty or contain only a single node?
How should subtree equality be determined? Should the node values and the tree structure be identical for two subtrees to be considered duplicates?
If multiple duplicate subtrees exist, should I return all of them, or just one representative subtree for each unique duplicate?
What is the expected output format? Specifically, should I return the root nodes of the duplicate subtrees, or some other representation?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

First, consider every single part of the tree as a potential subtree.
For each of these subtrees, create a description that captures its structure and the information it holds.
Now, take the first subtree and compare its description to the description of every other subtree.
If two subtrees have the exact same description, then we have found a duplicate.
Repeat this comparison process for every subtree against all other subtrees.
Keep a record of each unique duplicate subtree that you find to avoid reporting the same duplicate multiple times.

Code Implementation

def find_duplicate_subtrees_brute_force(root):
    duplicate_subtrees = []
    all_subtrees = []

    def traverse(node):
        if not node:
            return

        all_subtrees.append(node)
        traverse(node.left)
        traverse(node.right)

    traverse(root)

    def subtree_description(node):
        if not node:
            return "#"
        return str(node.val) + "," + subtree_description(node.left) + "," + subtree_description(node.right)

    for i in range(len(all_subtrees)):
        for j in range(i + 1, len(all_subtrees)):
            # Compare all possible subtrees.
            if subtree_description(all_subtrees[i]) == subtree_description(all_subtrees[j]):

                is_duplicate = False
                # Check if we've already recorded the duplicate.
                for previously_found in duplicate_subtrees:
                    if subtree_description(previously_found) == subtree_description(all_subtrees[i]):
                        is_duplicate = True
                        break

                if not is_duplicate:
                    # Record the duplicate if it's a new one.
                    duplicate_subtrees.append(all_subtrees[i])

    return duplicate_subtrees

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

First, create a way to turn each subtree into a string representation. This string should capture the structure and data of the entire subtree.
For any empty spot on the tree (like a missing branch), use a special marker in the string, so they are all consistently handled.
Now, go through the entire tree, turning each subtree into its string form. Store each unique string we encounter.
If we find a string that's already stored, that means we've found a duplicate subtree. Keep track of the root node of the duplicate subtree.
In the end, report all the root nodes of the duplicate subtrees we've found.

Code Implementation

def find_duplicate_subtrees(root):
    subtree_representation_map = {}
    duplicate_subtrees = []

    def get_subtree_string(node):
        if not node:
            return "#"

        # Construct a unique string for the subtree.
        subtree_string = str(node.val) + "," + get_subtree_string(node.left) + "," + get_subtree_string(node.right)

        # Check if the subtree string already exists.
        if subtree_string in subtree_representation_map:
            subtree_representation_map[subtree_string] += 1

            # Add the node to duplicates only once.
            if subtree_representation_map[subtree_string] == 2:

                # Mark this node as the root of a duplicate.
                duplicate_subtrees.append(node)
        else:

            # Store the string if it's the first time.
            subtree_representation_map[subtree_string] = 1

        return subtree_string

    get_subtree_string(root)

    # Return all duplicate subtree root nodes.
    return duplicate_subtrees

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty tree as input	Return an empty list immediately since there are no subtrees to compare.
A single-node tree	Return an empty list as a single node tree cannot have duplicate subtrees.
A tree where all nodes have the same value	The entire tree and all subtrees will be duplicates; ensure only the root of each distinct subtree structure is added to the result.
A tree where all nodes have distinct values	Return an empty list because no duplicate subtrees will exist.
Very large tree (potential stack overflow with recursion)	Consider using an iterative approach to traverse the tree instead of recursion, or increase stack size if using recursion.
Deeply skewed tree (e.g., a linked list)	This edge case tests the performance of the algorithm with imbalanced trees, potentially leading to worst-case time complexity; optimize the subtree comparison for such cases.
Trees with identical structure but different node values	The serialization/hashing method needs to incorporate node values to distinguish these trees.
Tree contains duplicate subtrees that are not direct children	The algorithm should identify all duplicate subtrees regardless of their level in the tree.