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.
1、后序遍历一遍树，同时计算子树和 (subtree sum)
2、用count 来记录 子树和 (subtree sum) 出现的次数
3、更新 max_count
4、把 count 和 子树和 (subtree sum) 加到 HashMap 里
public class Solution { Map<Integer, Integer> map = new HashMap<>(); int max_count = 0; public int[] findFrequentTreeSum(TreeNode root) { helper(root); Set<Map.Entry<Integer, Integer>> set = map.entrySet(); List<Integer> list = new ArrayList<>(); for (Map.Entry<Integer, Integer> item : set) { if (item.getValue() == max_count) list.add(item.getKey()); } int[] res = new int[list.size()]; for (int i = 0; i < list.size(); i++) { res[i] = list.get(i); } return res; } private int helper(TreeNode root) { if (root == null) return 0; int left_sum = helper(root.left); int right_sum = helper(root.right); int sum = left_sum + right_sum + root.val; int count = map.getOrDefault(sum, 0) + 1; max_count = Math.max(count, max_count); map.put(sum, count); return sum; } }
