<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*build(vector<int>pre,int prelo,int prehi,vector<int>in,int inlo,int inhi){
    if(prelo>prehi) return NULL;
    TreeNode*root=new TreeNode(pre[prelo]);
    if(prelo==prehi) return root;
    int i=inlo;
    while(i<inhi){
        if(in[i]==pre[prelo]) break;
        i++;
    }
    int leftcount=i-inlo;
    int rightcount=inhi-i;
    root->left=build(pre,prelo+1,prelo+leftcount,in,inlo,i-1);
    root->right=build(pre,prelo+leftcount+1,prehi,in,i+1,inhi);
    return root;

}
    TreeNode* buildTree(vector<int>& pre, vector<int>& in) {
        int n=pre.size();
        return build(pre,0,n-1,in,0,n-1);
        
    }
};
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