LeetCode 508: Most Frequent Subtree Sum Solution in Python – A Step-by-Step Guide

What Is LeetCode 508: Most Frequent Subtree Sum?

In LeetCode 508: Most Frequent Subtree Sum, you’re given the root of a binary tree, and your job is to find all subtree sums that occur most frequently. A subtree sum is the total of all node values in a subtree rooted at any node. For example, in a tree with root 5, left child 2, and right child -5, the subtree sums are 5 (root alone), 2, -5, and 2 (entire tree), with 2 appearing most often. This problem builds on LeetCode 501: Find Mode in Binary Search Tree, but applies to any binary tree.

Problem Statement

• Input: Root node of a binary tree (can be None).
• Output: List of integers—subtree sums with the highest frequency.
• Rules: Return all sums if multiple tie for most frequent; order doesn’t matter.

Constraints

• Number of nodes: 1 to 10⁴.
• Node values: -10⁵ to 10⁵.

Examples

• Input: [5,2,-5]
• Output: [2]
• Subtree sums: 5, 2, -5, 2 (5+2+-5) → 2 appears twice.
• Input: [5,2,-3]
• Output: [2,-3,4]
• Subtree sums: 5, 2, -3, 4 (5+2+-3) → all appear once.
• Input: [1]
• Output: [1]
• Single node, sum is 1.

Understanding the Problem: Summing Subtrees

To solve LeetCode 508: Most Frequent Subtree Sum in Python, we need a function that computes the sum of every subtree, counts their frequencies, and identifies the most common ones. A naive approach might compute sums repeatedly, but with up to 10⁴ nodes, we need efficiency. We’ll explore:

• Best Solution (Postorder with Hash Map): O(n) time, O(n) space—fast and smart.
• Alternative Solution (DFS with Array): O(n) time, O(n) space—simpler but less direct.

Let’s dive into the postorder solution with a friendly breakdown!

Best Solution: Postorder Traversal with Hash Map

Why Postorder with Hash Map Wins

The postorder traversal with a hash map is the best for LeetCode 508 because it’s efficient and intuitive. It computes each subtree sum once in O(n) time using a bottom-up approach (left, right, root), and tracks frequencies in a hash map with O(n) space. Finding the max frequency and corresponding sums is then a quick pass. It’s like tallying votes as you climb up the tree!

How It Works

Think of this as a tree-summing expedition:

• Step 1: Traverse Postorder:
• Recursively compute left and right subtree sums, then add the root.
• Step 2: Track Frequencies:
• Use a hash map to count how often each sum appears.
• Step 3: Find Modes:
• Identify the max frequency.
• Collect all sums with that frequency.
• Why It Works:
• Postorder ensures subtree sums are ready when needed.
• Hash map makes frequency counting instant.

It’s like a subtree sum census!

Step-by-Step Example

Example: [5,2,-5]

• Tree:
``` 5 / \ 2 -5 ```
• Init: count = {}, max_freq = 0.
• Postorder:
• Left (2): Sum = 2, count[2] = 1, max_freq = 1.
• Right (-5): Sum = -5, count[-5] = 1, max_freq = 1.
• Root (5): Sum = 5 + 2 + -5 = 2, count[2] = 2, max_freq = 2.
• Root alone: Sum = 5, count[5] = 1.
• Sums: 2 (twice), -5, 5.
• Find Modes: max_freq = 2, only 2 has freq 2.
• Result: [2].

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

Copy codeclass TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Solution:
    def findFrequentTreeSum(self, root: TreeNode) -> List[int]:
        if not root:
            return []

        # Step 1: Hash map for frequency
        count = {}

        # Step 2: Postorder traversal
        def postorder(node):
            if not node:
                return 0

            # Compute left and right sums
            left_sum = postorder(node.left)
            right_sum = postorder(node.right)

            # Total subtree sum
            total = node.val + left_sum + right_sum

            # Update frequency
            count[total] = count.get(total, 0) + 1

            return total

        postorder(root)

        # Step 3: Find max frequency
        max_freq = max(count.values())

        # Step 4: Collect sums with max frequency
        result = [sum_val for sum_val, freq in count.items() if freq == max_freq]

        return result

• Lines 10-11: Handle empty tree.
• Line 14: Hash map to store sum frequencies.
• Lines 17-26: Postorder helper:
• Base case: 0 for null.
• Recurse left and right.
• Compute total sum, update count.
• Return sum for parent.
• Line 29: Run traversal.
• Line 32: Find max frequency.
• Line 35: List comprehension for modes.
• Time Complexity: O(n)—visit each node once.
• Space Complexity: O(n)—hash map size.

It’s like a tree-sum vote counter!

Alternative Solution: Recursive DFS with Array Storage

Why an Alternative Approach?

The recursive DFS with array storage computes all subtree sums first, then counts frequencies. It’s O(n) time and O(n) space—simpler to grasp but requires two passes (collect sums, then find modes). Great for understanding tree recursion basics.

How It Works

Picture this as a two-stage tree hike:

• Step 1: Recursively collect all subtree sums in a list.
• Step 2: Count frequencies and find the most frequent sums.
• Step 3: Return the result.

It’s like gathering all sums, then tallying!

Step-by-Step Example

Example: [5,2,-3]

• Tree:
``` 5 / \ 2 -3 ```
• Step 1: Collect sums:
• Left: 2.
• Right: -3.
• Root: 5 + 2 + -3 = 4.
• Root alone: 5.
• sums = [2, -3, 4, 5].
• Step 2: Count frequencies:
• count = {2: 1, -3: 1, 4: 1, 5: 1}.
• Max freq = 1.
• Step 3: All sums have freq 1, result = [2, -3, 4, 5].
• Result: [2, -3, 4, 5].

Code for DFS Approach

Copy codeclass Solution:
    def findFrequentTreeSum(self, root: TreeNode) -> List[int]:
        if not root:
            return []

        # Step 1: Collect all subtree sums
        sums = []

        def dfs(node):
            if not node:
                return 0

            left_sum = dfs(node.left)
            right_sum = dfs(node.right)
            total = node.val + left_sum + right_sum
            sums.append(total)
            return total

        dfs(root)

        # Step 2: Count frequencies
        count = {}
        for s in sums:
            count[s] = count.get(s, 0) + 1

        # Step 3: Find max frequency and modes
        max_freq = max(count.values())
        result = [s for s, freq in count.items() if freq == max_freq]

        return result

• Time Complexity: O(n)—traversal plus linear pass.
• Space Complexity: O(n)—sums list and hash map.

It’s a sum-collecting trek!

Comparing the Two Solutions

• Postorder with Hash Map (Best):
• Pros: O(n), single pass, efficient.
• Cons: Hash map logic.
• DFS with Array (Alternative):
• Pros: O(n), simpler concept.
• Cons: Two passes, extra storage.

Postorder wins for elegance!

Additional Examples and Edge Cases

• [1]: [1].
• [3,-3]: [3, -3, 0].
• [0,0,0]: [0] (multiple zeros).

Postorder handles them all!

Complexity Recap

• Postorder: Time O(n), Space O(n).
• DFS: Time O(n), Space O(n).

Postorder’s the streamlined champ!

Key Takeaways

• Postorder: Bottom-up sums—learn at Python Basics!
• Hash Maps: Frequency counting made easy.
• Trees: Subtree fun.
• Python: Dicts shine!

Final Thoughts: Sum Those Subtrees!

LeetCode 508: Most Frequent Subtree Sum in Python is a delightful tree challenge. Postorder with a hash map is your fast track, while DFS with arrays keeps it basic. Want more? Try LeetCode 501: Find Mode in BST or LeetCode 652: Find Duplicate Subtrees. Ready to sum? Head to Solve LeetCode 508 on LeetCode and find those frequent sums today!