<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:
        int ls(int key, vector<int> &inorder)
        {
            int n = inorder.size();
            for (int i = 0; i < n; i++)
            {
                if (key == inorder[i]) return i;
            }
            return -1;
        }
    TreeNode* buildTree(vector<int> &preorder, vector<int> &inorder)
    {
        if (preorder.size() <= 0 || inorder.size() <= 0) return nullptr;
        TreeNode *root = new TreeNode(preorder[0]);
        preorder.erase(preorder.begin(), preorder.begin() + 1);
        if (inorder.size() == 1) return root;
        
        int mid = ls(root->val, inorder);

        vector<int> left;
        for(int i = 0;i<mid;i++) left.push_back(inorder[i]);
        root->left = buildTree(preorder, left);

        vector<int> right;
        for(int i = mid+1;i<inorder.size();i++) right.push_back(inorder[i]);
        root->right = buildTree(preorder, right);
        
        return root;
    }
};
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