<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).
# Definition for a binary tree node.
class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Solution:
    def findMode(self, root: TreeNode) -> List[int]:
        if not root:
            return []
        
        self.prev = None
        self.max_count = 0
        self.current_count = 0
        self.modes = []
        
        def inorder(node):
            if not node:
                return
            
            inorder(node.left)
            
            # Check if this is the first node or if the value has changed
            if self.prev == node.val:
                self.current_count += 1
            else:
                self.current_count = 1
            
            # Update modes if necessary
            if self.current_count > self.max_count:
                self.max_count = self.current_count
                self.modes = [node.val]
            elif self.current_count == self.max_count:
                self.modes.append(node.val)
            
            self.prev = node.val
            
            inorder(node.right)
        
        inorder(root)
        return self.modes

# Test the solution
# Constructing the tree:
#     1
#      \
#       2
#      /
#     2
root = TreeNode(1)
root.right = TreeNode(2)
root.right.left = TreeNode(2)

solution = Solution()
print(solution.findMode(root))  # Should print [2]
19800