Most Frequent Subtree Sum — LeetCode 508 Python Solution

MediumTreeDepth-First SearchHash TableBinary Tree
Problem
#508
Reading time
3 min

The problem

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.

Example

Input
root = [5,2,-3]
Output
[2,-3,4]

Python solution

Python
# Definition for a binary tree node.
# 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: Optional[TreeNode]) -> List[int]:
        def dfs(root: Optional[TreeNode]) -> int:
            if root is None:
                return 0
            l, r = dfs(root.left), dfs(root.right)
            s = l + r + root.val
            cnt[s] += 1
            return s

        cnt = Counter()
        dfs(root)
        mx = max(cnt.values())
        return [k for k, v in cnt.items() if v == mx]

Complexity

MeasureComplexity
TimeO(n)
SpaceO(n) auxiliary

Pattern: Tree Traversal

Choose the order — preorder, inorder, postorder, level — and the problem solves itself. LeetCode 508. Most Frequent Subtree Sum is filed here because LeetCode tags it Tree and Binary Tree, which is the vocabulary this hub collects.

The tree traversal guide has the Python template for the pattern and the 225 LeetCode problems that use it.

Related problems

Frequently asked questions

How hard is LeetCode 508. Most Frequent Subtree Sum?
LeetCode 508. Most Frequent Subtree Sum is rated Medium on LeetCode.
What is the time complexity of LeetCode 508. Most Frequent Subtree Sum?
The Python solution on this page runs in O(n).
What is the space complexity of LeetCode 508. Most Frequent Subtree Sum?
The Python solution on this page uses O(n) auxiliary space.
What topics does LeetCode 508. Most Frequent Subtree Sum cover?
LeetCode 508. Most Frequent Subtree Sum is tagged Tree, Depth-First Search, Hash Table and Binary Tree on LeetCode.

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