<p>Given the <code>root</code> of a binary search tree, return <em>a <strong>balanced</strong> binary search tree with the same node values</em>. If there is more than one answer, return <strong>any of them</strong>.</p>

<p>A binary search tree is <strong>balanced</strong> if the depth of the two subtrees of every node never differs by more than <code>1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/08/10/balance1-tree.jpg" style="width: 500px; height: 319px;" />
<pre>
<strong>Input:</strong> root = [1,null,2,null,3,null,4,null,null]
<strong>Output:</strong> [2,1,3,null,null,null,4]
<b>Explanation:</b> This is not the only correct answer, [3,1,4,null,2] is also correct.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/08/10/balanced2-tree.jpg" style="width: 224px; height: 145px;" />
<pre>
<strong>Input:</strong> root = [2,1,3]
<strong>Output:</strong> [2,1,3]
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>4</sup>]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 10<sup>5</sup></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 inorder_traversal(self, root):
        """
                5
               / \
              3   8
             / \   \
            2   4   10

           [2,3,4,5,8,10] -> if its a binary search tree is sorted
            otherwise no:

               1
                \
                2
                /
                3
            [1,3,2] 

        """

        result = []
        def inorder(node):
            if not node:
                return

            inorder(node.left)   
            result.append(node.val)
            inorder(node.right)

        inorder(root)
        return result

    def sorted_list_to_bst(self, nums):

        """
                [1, 2, 3, 4]
            [1,2]           [3,4]
            [1][2]          [3][4]
        """

        if not nums:
            return None
            
        mid = len(nums) // 2
        root = TreeNode(nums[mid])
        root.left = self.sorted_list_to_bst(nums[:mid])
        root.right = self.sorted_list_to_bst(nums[mid+1:])
        return root
    def sorted_list_to_bst(self, nums):

        """
                [1, 2, 3, 4]
            [1,2]           [3,4]
            [1][2]          [3][4]
        """

        if not nums:
            return None
            
        mid = len(nums) // 2
        root = TreeNode(nums[mid])
        root.left = self.sorted_list_to_bst(nums[:mid])
        root.right = self.sorted_list_to_bst(nums[mid+1:])
        return root

    def balanceBST(self, root: TreeNode) -> TreeNode:
        """
        1
         \
          2
           \
            3
        
          2
         / \
        1   3
             \
              4

        """
        inorder_traversal_list = self.inorder_traversal(root) # [1, 2, 3, 4]
        return self.sorted_list_to_bst(inorder_traversal_list)
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