从前序与中序遍历序列构造二叉树 C++

给定两个整数数组 preorder 和 inorder ，其中 preorder 是二叉树的先序遍历， inorder 是同一棵树的中序遍历，请构造二叉树并返回其根节点。

示例 1:

输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
输出: [3,9,20,null,null,15,7]

示例 2:

输入: preorder = [-1], inorder = [-1]
输出: [-1]

class Solution {
public:
    TreeNode* _buildTree(vector<int>& preorder, vector<int>& inorder,int& prei,int inbegin,int inend) 
    {
        if(inbegin>inend)
            return nullptr;
        TreeNode*root=new TreeNode(preorder[prei]);
        int rooti=inbegin;
        while(rooti<=inend)
        {
            if(inorder[rooti]==preorder[prei])
                break;
            else
                ++rooti;
        }
        if(inbegin<=rooti-1)
            root->left=_buildTree(preorder,inorder,++prei,inbegin,rooti-1);
        else
            root->left=nullptr;

        if(rooti+1<=inend)
            root->right=_buildTree(preorder,inorder,++prei,rooti+1,inend);
        else
            root->right=nullptr;
        return root;
    }
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) 
    {
        int prei=0;
        int inbegin=0,inend=inorder.size()-1;
        return _buildTree(preorder,inorder,prei,inbegin,inend);
    }
};