leetcode 501. Find Mode in Binary Search Tree

Given a binary search tree (BST) with duplicates, find all the mode(s) (the most frequently occurred element) in the given BST.

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].

Note: If a tree has more than one mode, you can return them in any order.

Follow up: Could you do that without using any extra space? (Assume that the implicit stack space incurred due to recursion does not count).

题目大意：求给定bst树种出现次数最多的数。
思路：中序遍历，记录先前结点的值和当前结点的值比较，并记录出现的次数。具体实现看代码。

/**
 * 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> ans;
    int Max = 0;
    void dfs(TreeNode* root, int& pre, int& cn) {
        if (root == nullptr) return ;
        dfs(root->left, pre, cn);
        if (root->val == pre) {
            cn++;
        } else if (pre != -1){
            if (cn > Max) {
                Max = cn;
                ans.clear();
                ans.emplace_back(pre);
                //cout << root->val << endl;
            } else if (cn == Max) {
                ans.emplace_back(pre);
            }
            cn = 1;
        }
        //cout << root->val << endl;
        pre = root->val;
        dfs(root->right, pre, cn);
    }
    vector<int> findMode(TreeNode* root) {
        int pre = -1;
        int cn = 1;
        if (root == nullptr) return ans;
        dfs(root, pre, cn);
        if (cn > Max) {
            ans.clear();
            ans.emplace_back(pre);
        } else if (cn == Max) {
            ans.emplace_back(pre);
        }
        return ans;
    }
};