問題：https://leetcode.com/problems/find-mode-in-binary-search-tree/?tab=Description
Given a binary search tree (BST) with duplicates, find all the mode(s) (the most frequently occurred element) in the given BST. 給一個二叉搜尋樹，求出現次數最多的數是哪個。
Assume a BST is defined as follows:
The left subtree of a node contains only nodes with keys less than or equal to the node’s key.
The right subtree of a node contains only nodes with keys greater than or equal to the node’s key.
Both the left and right subtrees must also be binary search trees.
For example:
Given BST [1,null,2,2],
1
\
2
/
2
return [2].
分析：

二叉搜尋樹的中序遍歷的結果恰好是所有數的遞增序列，根據中序遍歷結果，對於當前遍歷結點，標記maxCount為最大出現次數，tempCount為當前數字出現的次數，currentVal為當前儲存的值。
首先，tempCount++表示當前的數字出現次數+1，如果當前結點的值不等於儲存的值，就更新currentVal的值，並且將tempCount標記為1。
接下來，如果tempCount即當前數字出現的次數大於maxCount，就更新maxCount，並且將result陣列清零，並將當前數字放入result陣列中；如果tempCount只是等於maxCount，說明是出現次數一樣的，則將當前數字直接放入result陣列中。
參考C++程式碼：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> findMode(TreeNode* root) {
        inorder(root);
        return
 
 result;
    }
private:
   vector<int> result;
   int maxcount=0;
   int tem=0;
   int cur=0;
   void inorder(TreeNode* root){
       if(NULL==root) return;
       inorder(root->left);
       tem++;
       if(root->val!=cur){
           cur=root->val;
           tem=1;
       }
       if(tem>maxcount){
           maxcount=tem;
           result.clear();
           result.push_back(root->val);
       }
       else if(tem==maxcount){
           result.push_back(root->val);
       }
       inorder(root->right);
   }
};