二叉搜索树（1）

 使用中序遍历，返回第k个数来处理

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def kthLargest(self, root: TreeNode, k: int) -> int:
        def helper(root):
            return helper(root.left) + [root.val] + helper(root.right) if root else []
        return helper(root)[-k]

class Solution:
    def kthLargest(self, root: TreeNode, k: int) -> int:
        def dfs(root):
            if not root: return
            dfs(root.right)
            if self.k == 0: return
            self.k -= 1
            if self.k == 0: self.res = root.val
            dfs(root.left)
        self.k = k
        dfs(root)
        return self.res