Trim a binary search tree

Time: O(N); Space: O(H); easy

Given a binary search tree and the lowest and highest boundaries as L and R, trim the tree so that all its elements lies in [L, R] (R >= L).

You might need to change the root of the tree, so the result should return the new root of the trimmed binary search tree.

Example 1:

Input: root = {TreeNode} [1,0,2], L = 1, R = 2

  1
 / \
0   2

Output:

1
 \
  2

Example 2:

Input: root = {TreeNode} [3,0,4,None,2,None,None,None,None,1], L = 1, R = 3

  3
 / \
0   4
 \
  2
 /
1

Output:

    3
   /
  2
 /
1

[3]:

class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

[4]:

class Solution1(object):
    """
    Time: O(N)
    Space: O(H)
    """
    def trimBST(self, root, L, R):
        """
        :type root: TreeNode
        :type L: int
        :type R: int
        :rtype: do it in place
        """
        if not root:
            return None

        if root.val < L:
            return self.trimBST(root.right, L, R)

        if root.val > R:
            return self.trimBST(root.left, L, R)

        root.left, root.right = self.trimBST(root.left, L, R), self.trimBST(root.right, L, R)

        return root

[5]:

s = Solution1()

root = TreeNode(1)
root.left = TreeNode(0)
root.right = TreeNode(2)
L = 1
R = 2
s.trimBST(root, L, R)
assert root.val == 1
assert root.right.val == 2


root = TreeNode(3)
root.left = TreeNode(0)
root.right = TreeNode(4)
root.left.right = TreeNode(2)
root.left.right.left = TreeNode(1)
L = 1
R = 3
s.trimBST(root, L, R)
assert root.val == 3
assert root.left.val == 2
assert root.left.left.val == 1