501. Find Mode in Binary Search Tree -E

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).

A:

  ------------inorder BST traversal, save all values in the vector<int> and then count the mode ----------------

-------------------Method 2,  no extra space -----------------

/**
 * 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) {
        auto res = getModeAndCount(root);
        if(res[1] == 0)  // if the count is zero
        {
            return {};
        }else{
            res.pop_back();
            return res;
        }
    }
private:
    vector<int> getModeAndCount(TreeNode* root) // the last is the count of mode, and the rest are mode(s)
    { 
        if(not root)
            return {0,0};
        int rootCount = getCount(root, root->val);
        auto L = getModeAndCount(root->left);        
        auto R = getModeAndCount(root->right);
        if(L.back() == R.back())
        {
            L.pop_back();
            L.insert(L.end(), R.begin(), R.end());
        }else if (L.back() < R.back())   // use L to save the more result
        {
            L = R;
        }
        if(rootCount>L.back())
        {
            return {root->val, rootCount};
        }else if (rootCount == L.back()){
            L.back() = root->val;
            L.push_back(rootCount);
        }
        return L;
    }
              
    int getCount(TreeNode* root, int val)
    {
        if(not root)
            return 0;
        return (root->val == val? 1 : 0 ) + \
            (root->val >= val? getCount(root->left, val) : 0 ) + \
            (root->val <= val? getCount(root->right, val) : 0 );
    }
};

错误发生在 判断res[1] == 0 时，  一开始写成了 res.empty()