508. Most Frequent Subtree Sum-蒲公英云

508. Most Frequent Subtree Sum

Given the root of a binary tree, return the most frequent subtree sum. If there is a tie, return all the values with the highest frequency in any order.

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).

Example 1:

Input: root = [5,2,-3]
Output: [2,-3,4]

Example 2:

Input: root = [5,2,-5]
Output: [2]

题目：给定一个二叉树，找出所有出现次数最多的子树和的值

思路：DFS，用一个hashmap来保存所有字数和出现的次数，代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int getSum(TreeNode* node, unordered_map<int, int>& m, int& maxFre){
        if(!node) return 0;
        int left = getSum(node->left, m, maxFre);
        int right = getSum(node->right, m, maxFre);
        int sum = node->val + left + right;
        m[sum]++;
        if(m[sum] > maxFre) maxFre = m[sum];
        return sum;
    }
    vector<int> findFrequentTreeSum(TreeNode* root) {
        unordered_map<int, int> m;
        int maxFre = 0;
        getSum(root, m, maxFre);
        vector<int> res;
        for(auto mp : m){
            if(mp.second == maxFre)
                res.push_back(mp.first);
        }
        return res;
    }
};