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
SOLUTION:
# 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 sumOfTree(self, root):
if root:
curr = root.val + self.sumOfTree(root.left) + self.sumOfTree(root.right)
self.sums[curr] = self.sums.get(curr, 0) + 1
return curr
return 0
def findFrequentTreeSum(self, root: Optional[TreeNode]) -> List[int]:
self.sums = {}
self.sumOfTree(root)
mfreq = max(self.sums.values())
return [k for k, v in self.sums.items() if v == mfreq]
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