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

return [2, -3, 4], since all the values happen only once, return all of them in any order.

Examples 2

Input:

5 /  \2   -5

return [2], since 2 happens twice, however -5 only occur once.

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

题意：对于给定的一棵二叉树，找出现频率最高的子树之和。

思路：遍历和collections.Counter()计数器

 

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