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.

Examples 1
Input:

  5
 /  2   -3

Examples 2
Input:

  5
 /  2   -5

Note: You may assume the sum of values in any subtree is in the range of 32-bit signed integer.

題意：對於給定的一棵二叉樹，找出現頻率最高的子樹之和。

思路：遍歷和collections.Counter()計數器

# Definition for a binary tree node.
# class TreeNode():
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution():
    def findFrequentTreeSum(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        
 
"""
        cnt = collections.Counter()
        
        def sumOfSubTree(root):
            if not root:
                return 0
            return root.val + sumOfSubTree(root.left) + sumOfSubTree(root.right)
        
        def traverseTree(root):
            if not root:
                return None
            cnt[sumOfSubTree(root)] += 1
            traverseTree(root.left)
            traverseTree(root.right)
        traverseTree(root)
        max_sum = max(cnt.values() + [None])
        return [e for e, v in cnt.iteritems() if v == max_sum]

508. Most Frequent Subtree Sum (Medium)