題目

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.

難度

Medium

方法

對二叉樹進行后續(xù)遍歷，遞歸實現(xiàn)，將node節(jié)點及其子節(jié)點的和保存為node的val，最后就能獲取所有node及其子節(jié)點和。將各個node子節(jié)點和及其出現(xiàn)次數(shù)保存在dict中，最后選出出現(xiàn)次數(shù)最多的子節(jié)點和。

python代碼

import collections

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

class Solution(object):
    def findFrequentTreeSum(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root:
            return []
        self.counter = collections.Counter()
        self.postOrderTraverse(root)
        maxValue = max(self.counter.values())

        return [key for key in self.counter.keys() if self.counter[key] == maxValue]


    def postOrderTraverse(self, node):
        if node.left:
            node.val += self.postOrderTraverse(node.left)
        if node.right:
            node.val += self.postOrderTraverse(node.right)

        self.counter[node.val] += 1
        return node.val


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

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