https://programmercarl.com/0501.%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91%E4%B8%AD%E7%9A%84%E4%BC%97%E6%95%B0.html

视频讲解：https://www.bilibili.com/video/BV1fD4y117gp

1.这个题是我看完思路第一次一遍就成功ac的双指针的题，我觉得这个题的思路和前面几个题的思路都很相似，但是又有更好的思路，有代码层面的革新，让我觉得这个题我自己做了一遍之后，二刷的时候我一定要再来一遍，非递归的解法。

CPP

/**
 * 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:
    TreeNode* pre = NULL;
    int count = 0;
    int maxCount = 0;
    vector<int>res;
    void traversal(TreeNode* cur){
        if(cur == NULL)return ;

        traversal(cur->left);

        if(pre == NULL){
            count = 1;
        }else if(pre->val == cur->val){
            count++;
        }else{
            count = 1;
        }
        pre = cur;

        if(count == maxCount)res.push_back(cur->val);
        if(count > maxCount){
            maxCount = count;
            res.clear();
            res.push_back(cur->val);
        }

        traversal(cur->right);
    }
    vector<int> findMode(TreeNode* root) {
        traversal(root);
        return res;
    }
};