題目來源
查找二叉搜索樹種出現(xiàn)次數(shù)最多的節(jié)點(diǎn)。
我直接采用最簡單的方法，就是直接先把二叉樹中序遍歷。然后記錄下目前為止出現(xiàn)次數(shù)最多的節(jié)點(diǎn)。

/**
 * 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) {
        vector<int> nums;
        vector<int> res;
        dfs(root, nums);
        int m = 1, cur = 1;
        if (nums.size() == 0)
            return res;
        res.push_back(nums[0]);
        for (int i=1; i<nums.size(); i++) {
            if (nums[i-1] == nums[i])
                cur++;
            else
                cur = 1;
            if (cur == m)
                res.push_back(nums[i]);
            if (cur > m) {
                m = cur;
                res.clear();
                res.push_back(nums[i]);
            }
        }
        return res;
    }
    
    void dfs(TreeNode *node, vector<int> &nums)
    {
        if (!node)
            return;
        dfs(node->left, nums);
        nums.push_back(node->val);
        dfs(node->right, nums);
    }
};

然后搞了半天，怎么把vector搞掉，代碼如下：

/**
 * 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:
    TreeNode *pre = NULL;
    vector<int> findMode(TreeNode* root) {
        vector<int> res;
        int cur = 0, m = 1;
        dfs(root, res, m, cur);
        return res;
    }
    
    void dfs(TreeNode *node, vector<int> &res, int &m, int &cur)
    {
        if (!node)
            return;
        dfs(node->left, res, m, cur);
        if (pre && node->val == pre->val)
            cur++;
        else
            cur = 1;
        if (cur == m)
            res.push_back(node->val);
        if (cur > m) {
            m = cur;
            res.clear();
            res.push_back(node->val);
        }
        pre = node;
        dfs(node->right, res, m, cur);
    }
};