<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:
    void number_of_nodes(TreeNode* root,int &ans,vector<int>&v){
        if(root==NULL){
            return;
        }
        ans++;
        v.push_back(root->val);
        number_of_nodes(root->left,ans,v);
        number_of_nodes(root->right,ans,v);
    }
    
    TreeNode* createBalancedTreeWithValue1(int n) {
    if (n <= 0) {
        return NULL;
    }
    
    TreeNode* root = new TreeNode(1);
    queue<TreeNode*> q;
    q.push(root);

    int count = 1; 

    while (count < n) {
        TreeNode* current = q.front();
        q.pop();

        
        if (count < n) {
            current->left = new TreeNode(1);
            q.push(current->left);
            count++;
        }

        if (count < n) {
            current->right = new TreeNode(1);
            q.push(current->right);
            count++;
        }
    }

    return root;
}
void inorder(TreeNode* root,vector<int>v,int &index){
    if(root==NULL){
        return;
    }
    
    inorder(root->left,v,index);
    root->val=v[index];
    index++;
    inorder(root->right,v,index);
    
    
}
    TreeNode* balanceBST(TreeNode* root) {
        int ans=0;
        vector<int>v;
        number_of_nodes(root,ans,v);
        sort(v.begin(),v.end());
        TreeNode* newwroot= createBalancedTreeWithValue1(ans);

        int index=0;
        inorder(newwroot,v,index);
        return newwroot;


    }
};
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