human
Solve in Python:
S and T are strings composed of lowercase letters. In S, no letter occurs more than once.

S was sorted in some custom order previously. We want to permute the characters of T so that they match the order that S was sorted. More specifically, if x occurs before y in S, then x should occur before y in the returned string.

Return any permutation of T (as a string) that satisfies this property.

Example :
Input: 
S = "cba"
T = "abcd"
Output: "cbad"
Explanation: 
"a", "b", "c" appear in S, so the order of "a", "b", "c" should be "c", "b", and "a". 
Since "d" does not appear in S, it can be at any position in T. "dcba", "cdba", "cbda" are also valid outputs.

 

Note:

       S has length at most 26, and no character is repeated in S.
       T has length at most 200.
       S and T consist of lowercase letters only.
 Give me a the Python code you requested?
assistant
#Here is the Python code you requested.

# Definition for a binary tree node.
 # class TreeNode:
 #     def __init__(self, x):
 #         self.val = x
 #         self.left = None
 #         self.right = None
 
 class Solution:
     def splitBST(self, root,target):
         """
         :type root: TreeNode
         :type V: int
         :rtype: List[TreeNode]
         """
         
         def split_bst_recur(root, target):
             if not root:
                 return (None, None)
             
             if not root.left and not root.right:
                 if root.val <= target:
                     return (root, None)
                 else:
                     return(None, root)
             
             if root.val > target:
                 l, r = split_bst_recur(root.left, target)
                 root.left = r
                 return (l, root)
             else:
                 l, r = split_bst_recur(root.right, target)
                 root.right = l
                 return (root, r)
         
         if not root:
             return [[],[]]
         
         l, r = split_bst_recur(root, target)
         return [l, r]