Binary tree upside down

Time: O(N); Space: O(1); medium

Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. Return the new root.

Example 1:

Input: root = {TreeNode} [1,2,3,4,5]

    1
   / \
  2   3
 / \
4   5

Output: {TreeNode} [4,5,2,#,#,3,1]

  4
 / \
5   2
   / \
  3   1

Example 2:

Input: = {TreeNode} [1,2,3,4]

    1
   / \
  2   3
 /
4

Output: {TreeNode} [4,#,2,3,1]

4
 \
  2
 / \
3   1

[1]:

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

[2]:

class Solution1(object):
    """
    Time: O(N)
    Space: O(1)
    """
    def upsideDownBinaryTree(self, root):
        """
        :type root: TreeNode
        :rtype: root of the upside down tree
        """
        p, parent, parent_right = root, None, None

        while p:
            left = p.left
            p.left = parent_right
            parent_right = p.right
            p.right = parent
            parent = p
            p = left

        return parent

[6]:

s = Solution1()

root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(4)
root.left.right = TreeNode(5)
res = s.upsideDownBinaryTree(root)
assert res.val == 4
assert res.left.val == 5
assert res.right.val == 2
assert res.right.left.val == 3
assert res.right.right.val == 1

root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(4)
res = s.upsideDownBinaryTree(root)
assert res.val == 4
assert res.right.val == 2
assert res.right.left.val == 3
assert res.right.right.val == 1

[7]:

class Solution2(object):
    def upsideDownBinaryTree(self, root):
        """
        :type root: TreeNode
        :rtype: root of the upside down tree
        """
        return self.upsideDownBinaryTreeRecu(root, None)

    def upsideDownBinaryTreeRecu(self, p, parent):
        if p is None:
            return parent

        root = self.upsideDownBinaryTreeRecu(p.left, p)
        if parent:
            p.left = parent.right
        else:
            p.left = None
        p.right = parent

        return root

[8]:

s = Solution2()

root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(4)
root.left.right = TreeNode(5)
res = s.upsideDownBinaryTree(root)
assert res.val == 4
assert res.left.val == 5
assert res.right.val == 2
assert res.right.left.val == 3
assert res.right.right.val == 1

root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(4)
res = s.upsideDownBinaryTree(root)
assert res.val == 4
assert res.right.val == 2
assert res.right.left.val == 3
assert res.right.right.val == 1