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.
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?
本题和不用递归的方法中序遍历BST (Leetcode 94题) 基本上一样。我们只要在iteratively的过程中输出第K个val就可以了。
1 class Solution(object): 2 def kthSmallest(self, root, k): 3 """ 4 :type root: TreeNode 5 :type k: int 6 :rtype: int 7 """ 8 ans = [] 9 stack = [] 10 11 while stack or root: 12 while root: 13 stack.append(root) 14 root = root.left 15 root = stack.pop() 16 ans.append(root.val) 17 if len(ans) == k: 18 return ans[-1] 19 root = root.right 20 return ans[-1]
Leetcode 230. Kth Smallest Element in a BST
原文：http://www.cnblogs.com/lettuan/p/6250981.html