<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 insert(self, b_tree, new_node):
        curr_node = b_tree
        if curr_node is None:
            return new_node

        while True:
            if new_node.val > curr_node.val:
                if curr_node.right is None:
                    break
                curr_node = curr_node.right
            else:
                if curr_node.left is None:
                    break
                curr_node = curr_node.left
        if new_node.val > curr_node.val:
            curr_node.right = new_node
        else:
            curr_node.left = new_node
        return b_tree

    def balanceBST(self, root: TreeNode) -> TreeNode:
        # get a list of nodes sorted by inorder traversal
        nodes = []
        stack = []
        curr = root
        while curr or stack:
            while curr:
                stack.insert(0, curr)
                curr = curr.left

            curr = stack.pop(0)
            nodes.append(curr)

            curr = curr.right
            
        
        # remove arrows on the nodes
        for n in nodes:
            n.left = None
            n.right = None

        # insert always halved indexes,
        # so first the middle, then the middles of the middle, etc.
        new_tree = None
        queue = [(0, len(nodes))]

        while queue:
            start, end = queue.pop(0)
            if start >= end:
                continue
            middle = start + ((end - start) // 2)
            new_tree = self.insert(new_tree, nodes[middle])
            queue.append((start, middle))
            queue.append((middle+1, end))
        return new_tree



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