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.

Difficulty: interview

Starter code:

class Solution:
    def customSortString(self, S: str, T: str) -> str:
        

Solutions:
# 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]
             


# 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, V):
         """
         :type root: TreeNode
         :type V: int
         :rtype: List[TreeNode]
         """
         if not root:
             return [None,None]
         if root.val==V:
             temp= root.right
             root.right= None
             return [root, temp]
         if root.val<V:
             small, large= self.splitBST(root.right, V)
             root.right=small
             return [root, large]
         if root.val>V:
             small, large= self.splitBST(root.left, V)
             root.left=large
             return [small, root]
             
             
             

# 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, V):
         left_bucket = []
         right_bucket = []
         
         def explore(node, V):
             if not node:
                 return
             
             if node.val <= V:
                 left_bucket.append(node)
                 explore(node.right, V)
                 node.right = None
             else:
                 right_bucket.append(node)
                 explore(node.left, V)
                 node.left = None
         
         explore(root, V)
         
         for i in range(1, len(left_bucket)):
             parent = left_bucket[i-1]
             child = left_bucket[i]
             if child.val <= parent.val:
                 parent.left = child
             else:
                 parent.right = child
         
         for i in range(1, len(right_bucket)):
             parent = right_bucket[i-1]
             child = right_bucket[i]
             if child.val <= parent.val:
                 parent.left = child
             else:
                 parent.right = child 
                 
         left = None
         
         if len(left_bucket) > 0:
             left = left_bucket[0]
         right = None
         if len(right_bucket) > 0:
             right = right_bucket[0]
         
         return [left, right]
         
         """
         :type root: TreeNode
         :type V: int
         :rtype: List[TreeNode]
         """
         

# Given a Binary Search Tree (BST) with root node root, and a target value V, split the tree into two subtrees
 # where one subtree has nodes that are all smaller or equal to the target value, while the other subtree has all
 # nodes that are greater than the target value.  It's not necessarily the case that the tree contains a node with value V.
 #
 # Additionally, most of the structure of the original tree should remain.  Formally, for any child C with parent P in the original tree,
 # if they are both in the same subtree after the split, then node C should still have the parent P.
 #
 # You should output the root TreeNode of both subtrees after splitting, in any order.
 #
 # Example 1:
 #
 # Input: root = [4,2,6,1,3,5,7], V = 2
 # Output: [[2,1],[4,3,6,null,null,5,7]]
 # Explanation:
 # Note that root, output[0], and output[1] are TreeNode objects, not arrays.
 #
 # The given tree [4,2,6,1,3,5,7] is represented by the following diagram:
 #
 #           4
 #         /   \
 #       2      6
 #      / \    / \
 #     1   3  5   7
 #
 # while the diagrams for the outputs are:
 #
 #           4
 #         /   \
 #       3      6      and    2
 #             / \           /
 #            5   7         1
 # Note:
 #
 # The size of the BST will not exceed 50.
 # The BST is always valid and each node's value is different.
 
 # 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, V):
         """
         :type root: TreeNode
         :type V: int
         :rtype: List[TreeNode]
         """
 
         if not root:
             return [None, None]
 
         if root.val == V:
             tmp = root.right
             root.right = None   # 切记这里要把根节点的右节点置为None
             return [root, tmp]
         elif V < root.val:
             n1, n2 = self.splitBST(root.left, V)
             root.left = n2
             return [n1, root]
         else:
             n1, n2 = self.splitBST(root.right, V)
             root.right = n1
             return[root, n2]


# 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, V):
         """
         :type root: TreeNode
         :type V: int
         :rtype: List[TreeNode]
         """
         if not root:
             return [None, None]
         if root.val == V:
             a = root.right
             root.right = None
             return [root, a]
         elif root.val<V:
             small, large = self.splitBST(root.right, V)
             root.right = small
             return [root, large]
         else:
             small, large = self.splitBST(root.left, V)
             root.left = large
             return [root, small]