Given a binary search tree, write a function kthSmallest to find the kth smallest element in it.

Note: 
You may assume k is always valid, 1 ≤ k ≤ BST's total elements.

Example 1:

Input: root = [3,1,4,null,2], k = 13/ \1   4\2
Output: 1

Example 2:

Input: root = [5,3,6,2,4,null,null,1], k = 35/ \3   6/ \2   4/1
Output: 3

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?

二叉搜索树中序遍历呈现递增趋势

def kthSmallest(root,k):if not root:return Nonestack = []while stack or root:while root:stack.append(root)root = root.leftroot = stack.pop()k -= 1if k == 0:return root.valroot = root.right

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