508. 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.

题目含义：给一棵树，找出现次数最多的子树和。子树和就是一个结点以及它下方所有结点构成的子树的总和，在这些总和中找一个出现次数最多的结果，如果有很多个这样的结果，返回这些结果构成的数组

 1     private int maxCount = 0;
 2     private Map<Integer,Integer> sumMap = new HashMap<>();
 3     private int getNodeSum(TreeNode root) {
 4         if (root == null) return 0;
 5         int sum =  root.val + getNodeSum(root.left) +getNodeSum(root.right);
 6         int newCount = sumMap.getOrDefault(sum,0)+1;
 7         sumMap.put(sum,newCount);
 8         maxCount = Math.max(maxCount,newCount);
 9         return sum;
10     }
11     
12     public int[] findFrequentTreeSum(TreeNode root) {
13         getNodeSum(root);
14         List<Integer> sums = new ArrayList<>();
15         for (Map.Entry<Integer,Integer> entry:sumMap.entrySet())
16         {
17             if (entry.getValue() == maxCount)
18                 sums.add(entry.getKey());
19         }
20         int[] result = new int[sums.size()];
21         for (int i=0;i<sums.size();i++)
22         {
23             result[i] = sums.get(i);
24         }
25         return result;        
26     }