<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 balanceBST(self, root: TreeNode) -> TreeNode:
        
        all_nodes = []
        
        search = [root]
        while search:
            current = search.pop(-1)
            all_nodes.append(current)
            if current.left is not None:
                search.append(current.left)
            if current.right is not None:
                search.append(current.right)

                
        all_nodes.sort(key = lambda x: x.val)
        
        def create_sub_tree(l, r):
            w = r-l
            if w == 1:
                n = all_nodes[l]
                n.left = None
                n.right = None
                return n
            elif w == 0:
                return None
            
            m = (w-1)//2 + l
            n = all_nodes[m]
            n.left = create_sub_tree(l, m)
            n.right = create_sub_tree(m+1, r)
            return n
            
        return create_sub_tree(0, len(all_nodes))
            
20200