【LeetCode】508. Most Frequent Subtree Sum【M】【40】_鈥渢ree node鈥漮bject is not subscriptable-优快云博客

本文链接：https://blog.youkuaiyun.com/sscssz/article/details/56484002

本文介绍了一种使用递归算法解决寻找二叉树中最频繁出现的子树和的方法。通过遍历每个节点并计算其子树的和，再利用字典记录各子树和的出现频率，最终找出出现次数最多的子树和。

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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.

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这个用递归做还是很舒服的，而且就是标准的递归应用场景，对每个节点算sum，然后字典排序就行了

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

class Solution(object):
    def findFrequentTreeSum(self, root):
        if not root:
            return []
        dic = {}

        def calSum(root):
            if root == None:
                return 0
            s = calSum(root.left) + calSum(root.right) + root.val
            dic[s] = dic.get(s,0) + 1
            return s

        calSum(root)

        l = sorted(dic.iteritems(),key = lambda k:k[1],reverse = True)

        maxx = l[0][1]

        res = []
        for k,v in l:
            if v == maxx:
                res += k,
            else:
                break
        return res
        """
        :type root: TreeNode
        :rtype: List[int]
        """