LeetCode 230 Kth Smallest Element in a BST

this is actually pretty easy, we only needs to do in order traverse, and stopped at the k step

class Solution {
    public int kthSmallest(TreeNode root, int k) {
        if(root == null) return -1;
        int count = 0;
        
        LinkedList<TreeNode> stack = new LinkedList<>();
        TreeNode cur = root;
        //stack.push(cur); //we don't have to push cur in stack at this time because next while statement is ||
        while(!stack.isEmpty() || cur != null){
            while(cur != null){
                stack.push(cur);
                cur = cur.left;
            }
            cur = stack.pop();
            count = count + 1;
            if(count == k){
                return cur.val;
            }
            cur = cur.right;
        }
        return -1;//and even though it says: always assume k is valid, you have to consider this situation
    }
}

in the mean time, you also need to know how to solve it recursively:

class Solution {
  public ArrayList<Integer> inorder(TreeNode root, ArrayList<Integer> arr) {
    if (root == null) return arr;
    inorder(root.left, arr);
    arr.add(root.val);
    inorder(root.right, arr);
    return arr;
  }

  public int kthSmallest(TreeNode root, int k) {
    ArrayList<Integer> nums = inorder(root, new ArrayList<Integer>()); //just level order tranverse and get the k-1 th element.
    return nums.get(k - 1);
  }
}

Follow up:
What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?
https://leetcode.com/problems/kth-smallest-element-in-a-bst/solution/