<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>


 
class Solution {
public:
    void inorder(TreeNode* root, vector<int> & v){
        if(root==NULL) return ;
        inorder(root->left, v);
        v.push_back(root->val);
        inorder(root->right, v);
    }
    TreeNode* buildBST(vector<int> v, int s, int e){
        if(s>e) return NULL; 
        int mid = (s+e)/2;

        TreeNode* root = new TreeNode(v[mid]);
        if(s==e) return root; 

        root->left = buildBST(v,s, mid-1);
        root->right = buildBST(v,mid+1, e);
        return root; 


    }
    TreeNode* balanceBST(TreeNode* root) {
        vector<int> v; 
        inorder(root, v);

        return buildBST(v, 0, v.size()-1);
    }
};
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