Find Mode in Binary Search Tree
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0501.Find%20Mode%20in%20Binary%20Search%20Tree/images/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).
Tree; Depth-First Search; Binary Search Tree; Binary Tree
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func findMode(root *TreeNode) []int {
	mx, cnt := 0, 0
	var prev *TreeNode
	var ans []int
	var dfs func(root *TreeNode)
	dfs = func(root *TreeNode) {
		if root == nil {
			return
		}
		dfs(root.Left)
		if prev != nil && prev.Val == root.Val {
			cnt++
		} else {
			cnt = 1
		}
		if cnt > mx {
			ans = []int{root.Val}
			mx = cnt
		} else if cnt == mx {
			ans = append(ans, root.Val)
		}
		prev = root
		dfs(root.Right)
	}
	dfs(root)
	return ans
}