zoukankan      html  css  js  c++  java
  • Day4 -- Most Frequent Subtree Sum

    Most Frequent Subtree Sum



    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.

    solution

    #!/usr/bin/env python
    # -*- coding: utf-8 -*-
    # Time    : 2018/11/22 
    
    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]
            """
            map1 = {}
    
            def helper(root):
                if not root: return 0
                left_val = helper(root.left)
                right_val = helper(root.right)
                cur_sum = left_val + right_val + root.val
                print "cur_sum", cur_sum, left_val, right_val, root.val
                map1[cur_sum] = 1 if not map1.get(cur_sum) else map1[cur_sum] + 1
                return cur_sum
    
            helper(root)
            print "map", map1
            res = []
            value_list = [v for v in sorted(map1.values())]
            if value_list:
                max_fre = value_list[-1]
                res = [k for k, v in map1.items() if v == max_fre]
            return res
    
    
    if __name__ == '__main__':
        node1 = TreeNode(2)
        node1.left = TreeNode(-5)
        node1.right = TreeNode(1)
    
        node3 = TreeNode(7)
        node3.left = TreeNode(9)
        node3.right = TreeNode(-7)
    
        node2 = TreeNode(8)
        node2.left = node1
        node2.right = node3
    
        s = Solution()
        res = s.findFrequentTreeSum(node2)
        print res
    
    
  • 相关阅读:
    集合---Map
    一个机器部署多个tomcat
    JavaScript要不要加分号";"
    Nodejs 路径 /, ./, ../, ..// 的区别
    玩转Vue的24个小程序---基础篇
    如何创建Node.js Web服务器
    为什么Ajax XMLHttpRequest POST方法传递参数失败了
    字典元素如何遍历
    Beautiful Soup 如何获取到href
    如何查看Ajax请求
  • 原文地址:https://www.cnblogs.com/khal-Cgg/p/10019463.html
Copyright © 2011-2022 走看看