Introduction

In this blog post, we will explore how to solve the LeetCode problem "508" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given the root of a binary tree, return the most frequent subtree sum. If there is a tie, return all the values with the highest frequency in any order. The subtree sum of a node is defined as the sum of all the node values formed by the subtree rooted at that node (including the node itself). Example 1: Input: root = [5,2,-3] Output: [2,-3,4] Example 2: Input: root = [5,2,-5] Output: [2] Constraints: The number of nodes in the tree is in the range [1, 104]. -105 <= Node.val <= 105

Explanation

Here's the solution to find the most frequent subtree sum in a binary tree:

• Calculate Subtree Sums: Recursively traverse the tree, calculating the sum of each subtree. Store these sums in a hash map (dictionary) to track their frequencies.

• Track Frequency: Update the frequency count of each subtree sum in the hash map during the traversal.

• Find Most Frequent: After the traversal, identify the subtree sum(s) with the highest frequency and return them in a list.

• Time complexity: O(N), where N is the number of nodes in the tree, as each node is visited once. Storage Complexity: O(N) in the worst case (skewed tree) to store subtree sums and their frequencies.

Code

    from collections import defaultdict

class 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]:
        """
        Finds the most frequent subtree sum in a binary tree.

        Args:
            root: The root of the binary tree.

        Returns:
            A list of the subtree sums with the highest frequency.
        """

        subtree_sums = defaultdict(int)
        max_frequency = 0
        result = []

        def calculate_subtree_sum(node):
            """
            Recursively calculates the subtree sum for a given node.

            Args:
                node: The current node being processed.

            Returns:
                The sum of the subtree rooted at the current node.
            """
            if not node:
                return 0

            left_sum = calculate_subtree_sum(node.left)
            right_sum = calculate_subtree_sum(node.right)
            subtree_sum = node.val + left_sum + right_sum

            subtree_sums[subtree_sum] += 1
            return subtree_sum

        calculate_subtree_sum(root)

        for subtree_sum, frequency in subtree_sums.items():
            if frequency > max_frequency:
                max_frequency = frequency
                result = [subtree_sum]
            elif frequency == max_frequency:
                result.append(subtree_sum)

        return result