leetcode -- Kth Smallest Element in a BST -- 简单重点

https://leetcode.com/problems/kth-smallest-element-in-a-bst/

这里注意 BST的left subtree 所有**node都要小于root， right subtree 所有node都要大于**root。没有等于，并且是所有node

思路1 非递归用stack

用inorder的模板就行

class Solution(object):    def inorder(self, root, res, k):        stack = []        while stack or root:            if root:                stack.append(root)                root = root.left            else:                root = stack.pop()                res.append(root)                if len(res) == k:                    return root.val                root = root.right    def kthSmallest(self, root, k):        """        :type root: TreeNode        :type k: int        :rtype: int        """        return self.inorder(root, [], k)

思路2 递归

计算一个树的node个数，
参考http://www.rainatian.com/2015/07/06/leetcode-python-java-kth-smallest-element-in-a-bst/

class Solution(object):    def SizeOfTree(self, root):        if root is None:            return 0        return self.SizeOfTree(root.left) + self.SizeOfTree(root.right) + 1    def kthSmallest(self, root, k):        if root is None:            return 0        leftsize = self.SizeOfTree(root.left)        if leftsize == k-1:            return root.val        elif leftsize > k-1:            return self.kthSmallest(root.left,k)        else:            return self.kthSmallest(root.right,(k-1-leftsize))     

这个效率还没有非递归高，

思路3 O(BST的高度)

第二遍的时候再看

参考
http://bookshadow.com/weblog/2015/07/02/leetcode-kth-smallest-element-bst/