[leetcode]501. Find Mode in Binary Search Tree

题目链接：https://leetcode.com/problems/find-mode-in-binary-search-tree/#/description

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).

class Solution {public:    vector<int> findMode(TreeNode* root) {        put_map(root);        int maxtime=0;        vector<int> res;        //sort(m.begin(),m.end(),[&](pair<int,int> a,pair<int,int>b){return a.second<b.second;});        for(map<int,int>::iterator it=m.begin();it!=m.end();it++)        {            if(it->second>maxtime)            {                res.clear();                maxtime=it->second;                res.push_back(it->first);            }            else if(it->second==maxtime)            {                res.push_back(it->first);            }        }        return res;    }    void put_map(TreeNode* root)    {        if(!root)            return;        if(root->left) put_map(root->left);        m[root->val]+=1;        if(root->right) put_map(root->right);    };private:    map<int,int> m;};

class Solution {public:    vector<int> findMode(TreeNode* root) {        put_map(root);                return v;    }    void put_map(TreeNode* root)    {        if(!root)            return;        if(root->left) put_map(root->left);        currCount++;        if(root->val!=currValue)        {            currValue=root->val;            currCount=1;        }        if(currCount>maxCount)        {            maxCount=currCount;            v.clear();            v.push_back(root->val);        }        else if(currCount==maxCount)        {            v.push_back(root->val);        }        if(root->right) put_map(root->right);    };private:    vector<int> v;    int maxCount=0;    int currCount=0;    int currValue=0;};