<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:
    TreeNode* balanceBST(TreeNode* root) {
        // intput / condition: root
        // output / reference: nodes
        // variables:

        // problem space: f(balanced) = f(nodes) + f(in-order)
        vector<TreeNode*> nodes;
        dfs(root, nodes);
        
        return inOrderNodes(nodes);
    }

    TreeNode *inOrderNodesHelper(vector<TreeNode*> &nodes, int start, int end){
        // problem space: f(node) = f(left) + {root} + f(right)
        TreeNode *root = nullptr;
        int n = nodes.size();

        // exception
        if(start >= n || end >= n){
            return root;
        }

        // one node
        if(start == end){
            TreeNode *root = nodes[start];
            return root;
        }

        // two nodes
        if(start+1 == end){
            TreeNode *root = nodes[end];
            root->left = nodes[start];
            return root;
        }

        // general
        // three nodes

        int rootIndex = start + (end - start + 1)/2;
        root = nodes[rootIndex];

        root->left = inOrderNodesHelper(nodes, start, rootIndex-1);
        root->right = inOrderNodesHelper(nodes, rootIndex+1, end);

        return root;
    }

    TreeNode *inOrderNodes(vector<TreeNode*> &nodes){
        int n = nodes.size();
        return inOrderNodesHelper(nodes, 0, n-1);
    }

    void dfs(TreeNode *root, vector<TreeNode*> &nodes){
        if(root == nullptr){
            return;
        }
        
        dfs(root->left, nodes);

        nodes.push_back(new TreeNode(root->val));

        dfs(root->right, nodes);
    }
};
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