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# Kth Smallest Element in a BST LeetCode Solution

## Problem – Kth Smallest Element in a BST

Given the `root` of a binary search tree, and an integer `k`, return the `kth` smallest value (1-indexed) of all the values of the nodes in the tree.

Example 1:

``````Input: root = [3,1,4,null,2], k = 1
Output: 1``````

Example 2:

``````Input: root = [5,3,6,2,4,null,null,1], k = 3
Output: 3``````

Constraints:

• The number of nodes in the tree is `n`.
• `1 <= k <= n <= 104`
• `0 <= Node.val <= 104`

Follow up: 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?

### Kth Smallest Element in a BST LeetCode Solution in Java

``````def findNode(node, res):
if len(res) > 1:
return

if node.left:
findNode(node.left, res)

res -= 1
if res == 0:
res.append(node.val)
return

if node.right:
findNode(node.right, res)

res = [k]
findNode(root, res)
return res
``````

### Kth Smallest Element in a BST LeetCode Solution in Python

``````def kthSmallest(self, root, k):
self.k = k
self.res = None
self.helper(root)
return self.res

def helper(self, node):
if not node:
return
self.helper(node.left)
self.k -= 1
if self.k == 0:
self.res = node.val
return
self.helper(node.right)``````

### Kth Smallest Element in a BST LeetCode Solution in c++

``````int kthSmallest(TreeNode* root, int k) {
return find(root, k);
}
int find(TreeNode* root, int& k) {
if (root) {
int x = find(root->left, k);
return !k ? x : !--k ? root->val : find(root->right, k);
}
}
``````
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