# [Solved] 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.

## Question

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`

## Python 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 make_list(self,root,l):
if root == None:
return

self.make_list(root.left,l)
l.append(root.val)
self.make_list(root.right,l)

def increasingBST(self, root: TreeNode) -> TreeNode:
l = []
self.make_list(root,l)
ans = TreeNode(0)
curr = ans

for i in l:
curr.left = None
curr.right = TreeNode(i)
curr = curr.right

return ans.right```