<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:
  vector<int> findMode(TreeNode* root) {
    vector<int> ans;
    int count = 0;
    int maxCount = 0;

    inorder(root, count, maxCount, ans);
    return ans;
  }

 private:
  TreeNode* pred = nullptr;

  void inorder(TreeNode* root, int& count, int& maxCount, vector<int>& ans) {
    if (root == nullptr)
      return;

    inorder(root->left, count, maxCount, ans);
    updateCount(root, count, maxCount, ans);
    inorder(root->right, count, maxCount, ans);
  }

  void updateCount(TreeNode* root, int& count, int& maxCount,
                   vector<int>& ans) {
    if (pred && pred->val == root->val)
      ++count;
    else
      count = 1;

    if (count > maxCount) {
      maxCount = count;
      ans = {root->val};
    } else if (count == maxCount) {
      ans.push_back(root->val);
    }

    pred = root;
  }
};
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