Increasing Order Search Tree

Given the root of a binary search tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only one right child.

Example 1:

Input: root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]

Example 2:

Input: root = [5,1,7]
Output: [1,null,5,null,7]

Constraints:

• The number of nodes in the given tree will be in the range [1, 100].
• 0 <= Node.val <= 1000

SOLUTION:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def insert(self, root, key):
        if root:
            if key <= root.val:
                root.left = self.insert(root.left, key)
            else:
                root.right = self.insert(root.right, key)
        else:
            root = TreeNode(val = key)
        return root

    def insort(self, root):
        if root:
            self.insort(root.left)
            self.newroot = self.insert(self.newroot, root.val)
            self.insort(root.right)

    def increasingBST(self, root: TreeNode) -> TreeNode:
        self.newroot = None
        self.insort(root)
        return self.newroot