<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>

/**
 * 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* buildTree(vector<int>& preorder, vector<int>& inorder) {
        
        int n = preorder.size();

        // Base Case
        if (n == 0) {
            return nullptr;
        }

        int root_val = preorder[0];
        int root_val_inorder_index = findIndex(root_val, inorder);

        int left_subtree_size = root_val_inorder_index;
        int right_subtree_size = n - root_val_inorder_index - 1;
        
        vector<int> left_subtree_preorder;
        vector<int> left_subtree_inorder;
        for (int i = 1; i < left_subtree_size + 1; ++i) {
            left_subtree_preorder.push_back(preorder[i]);
        }
        for (int i = 0; i < left_subtree_size; ++i) {
            left_subtree_inorder.push_back(inorder[i]);
        }

        vector<int> right_subtree_preorder;
        vector<int> right_subtree_inorder;
        for (int i = 1 + left_subtree_size; i < 1 + left_subtree_size + right_subtree_size; ++i) {
            right_subtree_preorder.push_back(preorder[i]);
        }
        for (int i = root_val_inorder_index + 1; i < root_val_inorder_index + 1 + right_subtree_size; ++i) {
            right_subtree_inorder.push_back(inorder[i]);
        }

        TreeNode* root = new TreeNode(root_val);
        root->left = this->buildTree(left_subtree_preorder, left_subtree_inorder);
        root->right = this->buildTree(right_subtree_preorder, right_subtree_inorder);

        return root;
    }

    int findIndex(int x, vector<int> tree_vector) {
        for (int i = 0; i < tree_vector.size(); ++i) {
            if (tree_vector[i] == x) {
                return i;
            }
        }
        return -1;
    }

};
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