<p>Given the <code>root</code> of a binary search tree (BST) with duplicates, return <em>all the <a href="https://en.wikipedia.org/wiki/Mode_(statistics)" target="_blank">mode(s)</a> (i.e., the most frequently occurred element) in it</em>.</p>

<p>If the tree has more than one mode, return them in <strong>any order</strong>.</p>

<p>Assume a BST is defined as follows:</p>

<ul>
	<li>The left subtree of a node contains only nodes with keys <strong>less than or equal to</strong> the node&#39;s key.</li>
	<li>The right subtree of a node contains only nodes with keys <strong>greater than or equal to</strong> the node&#39;s key.</li>
	<li>Both the left and right subtrees must also be binary search trees.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/11/mode-tree.jpg" style="width: 142px; height: 222px;" />
<pre>
<strong>Input:</strong> root = [1,null,2,2]
<strong>Output:</strong> [2]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [0]
<strong>Output:</strong> [0]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>4</sup>]</code>.</li>
	<li><code>-10<sup>5</sup> &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> 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:
    void inor(TreeNode* root, vector<int>& vec) {
        if (!root) return;
        inor(root->left, vec);
        vec.push_back(root->val);
        inor(root->right, vec);
    }
    
    vector<int> findMode(TreeNode* root) {
        vector<int> v;
        inor(root, v);
        
        if (v.empty()) return {};
        
        vector<pair<int, int>> freq;
        int count = 1;
        
        for (int i = 1; i < v.size(); i++) {
            if (v[i] == v[i-1]) {
                count++;
            } else {
                freq.push_back({v[i-1], count});
                count = 1;
            }
        }
        freq.push_back({v.back(), count});
        
        sort(freq.begin(), freq.end(), [](const pair<int, int>& a, const pair<int, int>& b) {
            return a.second > b.second;
        });
        
        vector<int> result;
        int maxFreq = freq[0].second;
        
        for (const auto& p : freq) {
            if (p.second == maxFreq) {
                result.push_back(p.first);
            } else {
                break;
            }
        }
        
        return result;
    }
};
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