Leetcode-508. Most Frequent Subtree Sum

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

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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.

这个题目也挺简单的，采用后序遍历，然后把所有的结果存到hashmap里面，同时用一个临时变量记录最大的次数是多少，最后遍历map就可以了。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private Map sum2count = new HashMap();
    private int max_count = 0;
    public int[] findFrequentTreeSum(TreeNode root) {
        help(root);
        Set> set = sum2count.entrySet();
        List results = new ArrayList();
        for(Map.Entry item : set){
            if(item.getValue() == max_count) results.add(item.getKey());
        }
        int[] returns = new int[results.size()];
        for(int i = 0; i < results.size(); i ++){
            returns[i] = results.get(i);
        }
        return returns;
    }

    private int help(TreeNode root){
        if(root == null) return 0;
        int left_sum = help(root.left);
        int right_sum = help(root.right);
        int sum = left_sum + right_sum + root.val;
        int count = sum2count.getOrDefault(sum, 0) + 1;
        max_count = count > max_count ? count : max_count;
        sum2count.put(sum, count);
        return sum;
    }
}