Most Frequent Subtree Sum

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]

Constraints:

• The number of nodes in the tree is in the range [1, 104].
• -105 <= Node.val <= 105

Source: https://leetcode.com/problems/most-frequent-subtree-sum/

Leetcode Difficulty: Medium

Code:

class TreeNode(object):
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Solution(object):
    
    def __init__(self):
        self.subtree_sums = {}
        self.max_freq=-1
        
    def get_max_subtree_sum(self, root):
        if root==None: 
            return 0
        
        leftsum = self.get_max_subtree_sum(root.left)
        rightsum = self.get_max_subtree_sum(root.right)
        currval = root.val
        
        summation = currval + rightsum + leftsum
        self.subtree_sums[summation] = self.subtree_sums.get(summation, 0)+1
        self.max_freq = max(self.max_freq, self.subtree_sums[summation])
        return summation
        
    def findFrequentTreeSum(self, root):
        self.get_max_subtree_sum(root)
        result=[]
        for k,v in self.subtree_sums.items():
            if v==self.max_freq:
                result.append(k)
        return result

Thought Process / Explanation:

We know that sum of the subtree rooted at a certain node is left sum + right sum + curr val. We need to store the sum with it's count and later return only max frequency sum -- for this keeping a dictionary sounds good and also maintaining global max_freq counter for comparison.

Thank You!