<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.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> ans;
    TreeNode* sortedArrayToBST(vector<int> nodes) {
        if(nodes.size() == 0) return NULL;
        if(nodes.size() == 1) return new TreeNode(nodes[0]);
        int mid = nodes.size() / 2;
        TreeNode* currRoot = new TreeNode(nodes[mid]);
        vector<int> leftHalf(nodes.begin(), nodes.begin() + mid);
        vector<int> rightHalf(nodes.begin() + mid + 1, nodes.end());
        currRoot->left = sortedArrayToBST(leftHalf);
        currRoot->right = sortedArrayToBST(rightHalf);
        return currRoot;
    }

    void inOrderTraversal(TreeNode* root) {
        if(!root) return;
        inOrderTraversal(root->left);
        ans.push_back(root->val);
        inOrderTraversal(root->right);
    }
    TreeNode* balanceBST(TreeNode* root) {
        inOrderTraversal(root);
        return sortedArrayToBST(ans);
    }
};
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