Most frequent subtree sum

Time: O(N); Space: O(N); medium

Given the root of a tree, you are asked to find the most frequent subtree sum.

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).

So what is the most frequent subtree sum value? If there is a tie, return all the values with the highest frequency in any order.

Example 1:

  5
 / \
2  -3

Input: root = {TreeNode} [5,2,-3]

Output: [2,-3,4]

Explanation:

  • All the values happen only once, return all of them in any order.

Example 2:

  5
 / \
2  -5

Input: root = {TreeNode} [5,2,-5]

Output: [2]

Explanation:

  • 2 happens twice, however -5 only occur once.

[7]:

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

[8]:

import collections

class Solution1(object):
    def findFrequentTreeSum(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        def countSubtreeSumHelper(root, counts):
            if not root:
                return 0
            total = root.val + \
                    countSubtreeSumHelper(root.left, counts) + \
                    countSubtreeSumHelper(root.right, counts)
            counts[total] += 1
            return total

        counts = collections.defaultdict(int)
        countSubtreeSumHelper(root, counts)
        max_count = max(counts.values()) if counts else 0

        return [total for total, count in counts.items()if count == max_count]

[9]:

s = Solution1()

root = TreeNode(5)
root.left = TreeNode(2)
root.right = TreeNode(-3)
assert s.findFrequentTreeSum(root) == [2,-3,4]

root = TreeNode(5)
root.left = TreeNode(2)
root.right = TreeNode(-5)
assert s.findFrequentTreeSum(root) == [2]