508. Most Frequent Subtree Sum ----------M

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]

 

Constraints:

• The number of nodes in the tree is in the range [1, 104].
• -105 <= Node.val <= 105

A:

就是递归调用，并把结果存在 map里

/**
 * 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:
    vector<int> findFrequentTreeSum(TreeNode* root) {
        unordered_map<int,int> M;
        helper(root, M);
        int maxCount = 0;
        for(const auto p : M){
            maxCount = max(maxCount, p.second);
        }
        vector<int> res;
        for(const auto p : M){
            if(p.second == maxCount){
                res.push_back(p.first);
            }
        }
        return res;
    }
private:
    int helper(TreeNode* root, unordered_map<int,int> & M){
        if(!root)
            return 0;
        int l = helper(root->left, M);
        int r = helper(root->right, M);
        int sum = l + r + root->val;
        M[sum] +=1;
        return sum;
    }
};