Leetcode 508. Most Frequent Subtree Sum (二叉树后序遍历好题)-Toy模板网

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Most Frequent Subtree Sum
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Medium
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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

解法1：用后序遍历！文章来源地址https://www.toymoban.com/news/detail-836987.html

/**
 * 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) {
        helper(root);
        vector<int> res;        
        for (auto elem: um) {
            if (maxFreq == elem.second) {
                res.push_back(elem.first);
            }
        }
        return res;
    }
private:
    map<int, int> um; //<sum, freq>
    int maxFreq = 0;
    int helper(TreeNode* root) {
        if (!root) return 0;
        int left = helper(root->left);
        int right = helper(root->right);
        int sum = left + right + root->val;

        um[sum]++;
        if (maxFreq < um[sum]) {
            maxFreq = um[sum];
        }
        return sum;
    }
};