<p>Given the <code>root</code> of a binary search tree, rearrange the tree in <strong>in-order</strong> 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.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/17/ex1.jpg" style="width: 600px; height: 350px;" />
<pre>
<strong>Input:</strong> root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
<strong>Output:</strong> [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/17/ex2.jpg" style="width: 300px; height: 114px;" />
<pre>
<strong>Input:</strong> root = [5,1,7]
<strong>Output:</strong> [1,null,5,null,7]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the given tree will be in the range <code>[1, 100]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 1000</code></li>
</ul>

# 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 increasingBST(self, root: TreeNode) -> TreeNode:

        ptr = TreeNode()
        res = ptr

        def inorder(root):
            nonlocal ptr
            if root:
                inorder(root.left)
                ptr.right = TreeNode(root.val)
                ptr = ptr.right
                inorder(root.right)
        
        inorder(root)
        return res.right
        
        

        

        
16500