<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 {
    TreeNode* construct(vector<int>preorder, vector<int>inorder){
        if(preorder.size()==0){
            return nullptr;
        }
        else if(preorder.size()==1){
            TreeNode *root = new TreeNode(preorder[0]);
            return root;
        }
        else if(preorder.size()==2){
            TreeNode *root = new TreeNode(preorder[0]);
            TreeNode *children = new TreeNode(preorder[1]);
            if(preorder[1]==inorder[1]){
                root->right = children; 
            }
            else{
                root->left = children;
            }
            return root;
        }
        // else if(preorder.size()==3){
        //     TreeNode* root = new TreeNode(preorder[0]);
        //     root->left = new TreeNode(preorder[1]);
        //     root->right = new TreeNode(preorder[2]);
        //     return root;
        // }

        int rootVal = preorder[0];
        vector<int>leftIn, leftPre, rightIn, rightPre;

        int findR = 0;
        for(int i = 0 ; i < inorder.size() ; i++){
            if(rootVal == inorder[i]){
                findR = 1;
                continue;
            }
            if(findR){
                rightIn.push_back(inorder[i]);
            }
            else{
                leftIn.push_back(inorder[i]);
            }

        }
        int left = leftIn.size();
        for(int i = 1 ; i < preorder.size() ; i++){
            if(i<=left){
                leftPre.push_back(preorder[i]);
            }   
            else{
                rightPre.push_back(preorder[i]);
            }
        }

        TreeNode* root = new TreeNode(rootVal);
        root->left = construct(leftPre, leftIn);
        root->right = construct(rightPre, rightIn);

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

        return construct(preorder,inorder);
        
    }
};
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