<p>Given two integer arrays <code>preorder</code> and <code>inorder</code> where <code>preorder</code> is the preorder traversal of a binary tree and <code>inorder</code> is the inorder traversal of the same tree, construct and return <em>the binary tree</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/tree.jpg" style="width: 277px; height: 302px;" />
<pre>
<strong>Input:</strong> preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
<strong>Output:</strong> [3,9,20,null,null,15,7]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> preorder = [-1], inorder = [-1]
<strong>Output:</strong> [-1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= preorder.length &lt;= 3000</code></li>
	<li><code>inorder.length == preorder.length</code></li>
	<li><code>-3000 &lt;= preorder[i], inorder[i] &lt;= 3000</code></li>
	<li><code>preorder</code> and <code>inorder</code> consist of <strong>unique</strong> values.</li>
	<li>Each value of <code>inorder</code> also appears in <code>preorder</code>.</li>
	<li><code>preorder</code> is <strong>guaranteed</strong> to be the preorder traversal of the tree.</li>
	<li><code>inorder</code> is <strong>guaranteed</strong> to be the inorder traversal of the tree.</li>
</ul>

class Solution {
public:

    TreeNode* helper(vector<int> preorder, vector<int> inorder)
    {
        if (preorder.size() == 0) return nullptr;

        TreeNode* root = new TreeNode(preorder[0]);
        int idx = -1;

        for (int i = 0; i < inorder.size(); i++)
        {
            if (root->val == inorder[i])
            {
                idx = i;
                break;
            }
        }

        vector<int> lpreorder;
        vector<int> linorder;

        for (int i = 0; i < idx; i++)
        {
            lpreorder.push_back(preorder[i+1]);
            linorder.push_back(inorder[i]);
        }

        vector<int> rpreorder;
        vector<int> rinorder;

        for (int i = idx+1; i < preorder.size(); i++)
        {
            rpreorder.push_back(preorder[i]);
            rinorder.push_back(inorder[i]);
        }

        TreeNode* left = helper(lpreorder, linorder); 
        TreeNode* right = helper(rpreorder, rinorder); 
        root->left = left;
        root->right = right;

        return root;
    }

    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        TreeNode* root = helper(preorder, inorder);
        return root;
    }
};
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