题目描述：

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],

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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).

使用Hash表可以快速解决。

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/**
 * 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 {
    vector<int> ans;
    unordered_map<int, int> val;
    int maxOccur = 0;
public:
    vector<int> findMode(TreeNode* root) {
        inOrder(root);
        for (auto i : val) {
            if (i.second == maxOccur) 
                ans.push_back(i.first);
        }
        return ans;
    }
    
    void inOrder(TreeNode *node) {
        if(!node) return ;
        inOrder(node->left);
        maxOccur = max(maxOccur, ++val[node->val]);
        inOrder(node->right);
    }
};

在要求不使用额外存储空间的情况下，因为这是一棵BST，所以中序遍历结果是一个有序序列，问题就转化为在一个有序序列中连续出现次数最多的元素有哪些。除了维护结果集以外，还要维护结果集中元素的出现次数，上一个元素的值和当前元素的出现次数。当前元素的出现次数超过结果集中元素的出现次数后就清空结果集，然后增加当前元素。

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/**
 * 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 {
    vector<int> ans;
    int lastElem = INT_MAX;
    int lastElemOccurTimes;
    int valInAnsOccurTimes = 0;
public:
    vector<int> findMode(TreeNode* root) {
        inOrder(root);
        return ans;
    }
    
    void inOrder(TreeNode *node) {
        if(!node) return ;
        inOrder(node->left);
        if (lastElem == node->val) {
            lastElemOccurTimes++;
        }
        else {
            lastElemOccurTimes = 1;
            lastElem = node->val;
        }
        
        if (lastElemOccurTimes == valInAnsOccurTimes) {
            ans.push_back(node->val);
        }
        else if (lastElemOccurTimes > valInAnsOccurTimes) {
            valInAnsOccurTimes = lastElemOccurTimes;
            ans.clear();
            ans.push_back(node->val);
        }
            
        inOrder(node->right);
    }
};