# 230-kth-smallest-element-in-a-bst


Try it on <a href='https://leetcode.com/problems/230-kth-smallest-element-in-a-bst'>leetcode</a>

## Description
<div class="description">
<div><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>Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/28/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>Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/28/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>
</div>
</div>

## Solution(Python)
```Python
class Solution:
    def kthSmallest(self, root, k):
        """
        :type root: TreeNode
        :type k: int
        :rtype: int
        """

        cur = root

        while cur:
            if cur.left is None:
                k -= 1
                if k == 0:
                    return cur.val
                cur = cur.right
            else:
                pre = cur.left

                while pre.right and pre.right is not cur:
                    pre = pre.right

                if pre.right is None:
                    pre.right = cur
                    cur = cur.left
                else:
                    pre.right = None
                    k -= 1
                    if k == 0:
                        return cur.val
                    cur = cur.right

```