Kth Smallest Element in a BST
<p>Given the <code>root</code> of a binary search tree, and an integer <code>k</code>, return <em>the</em> <code>k<sup>th</sup></code> <em>smallest value (<strong>1-indexed</strong>) of all the values of the nodes in the tree</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0230.Kth%20Smallest%20Element%20in%20a%20BST/images/kthtree1.jpg" style="width: 212px; height: 301px;" />
<pre>
<strong>Input:</strong> root = [3,1,4,null,2], k = 1
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0230.Kth%20Smallest%20Element%20in%20a%20BST/images/kthtree2.jpg" style="width: 382px; height: 302px;" />
<pre>
<strong>Input:</strong> root = [5,3,6,2,4,null,null,1], k = 3
<strong>Output:</strong> 3
</pre>

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

<ul>
	<li>The number of nodes in the tree is <code>n</code>.</li>
	<li><code>1 &lt;= k &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= Node.val &lt;= 10<sup>4</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> If the BST is modified often (i.e., we can do insert and delete operations) and you need to find the kth smallest frequently, how would you optimize?</p>
Tree; Depth-First Search; Binary Search Tree; Binary Tree
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int kthSmallest(TreeNode root, int k) {
        Deque<TreeNode> stk = new ArrayDeque<>();
        while (root != null || !stk.isEmpty()) {
            if (root != null) {
                stk.push(root);
                root = root.left;
            } else {
                root = stk.pop();
                if (--k == 0) {
                    return root.val;
                }
                root = root.right;
            }
        }
        return 0;
    }
}