501. 二叉搜索树中的众数

/**
 * 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>a;
    map<int,int>mp;
    void order(TreeNode *t){
        if(t==NULL){
            return;
        }else{
            order(t->left);
            mp[t->val]++;
            order(t->right);
        }
    }
    set<int>st;
    vector<int> findMode(TreeNode* root) {
        order(root);
        int k=0;
        for(auto t:mp){
            if(t.second>=k){
                k=t.second;
            }
        }
        for(auto t:mp){
            if(t.second==k){
                a.push_back(t.first);
            }
        }
        return a;
    }
};

class Solution {
private:
 
void searchBST(TreeNode* cur, unordered_map<int, int>& map) { // 前序遍历
    if (cur == NULL) return ;
    map[cur->val]++; // 统计元素频率
    searchBST(cur->left, map);
    searchBST(cur->right, map);
    return ;
}
bool static cmp (const pair<int, int>& a, const pair<int, int>& b) {
    return a.second > b.second;
}
public:
    vector<int> findMode(TreeNode* root) {
        unordered_map<int, int> map; // key:元素，value:出现频率
        vector<int> result;
        if (root == NULL) return result;
        searchBST(root, map);
        vectorint, int>> vec(map.begin(), map.end());
        sort(vec.begin(), vec.end(), cmp); // 给频率排个序
        result.push_back(vec[0].first);
        for (int i = 1; i < vec.size(); i++) {
            // 取最高的放到result数组中
            if (vec[i].second == vec[0].second) result.push_back(vec[i].first);
            else break;
        }
        return result;
    }
};