[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