Construct Binary Tree from Preorder and Inorder Traversal [Leetcode]

Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.

Solution: the same idea as the solution to Construct Binary Tree from Inorder and Postorder Traversal.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(inorder.size()!=preorder.size() || inorder.size()==0)
            return NULL;
        int curVal = preorder[0];
        int i=0;
        while(i<inorder.size()){
            if(inorder[i]==curVal)
                break;
            ++i;
        }
        
        vector<int> inR(inorder.begin()+i+1, inorder.end());
        inorder.erase(inorder.begin()+i, inorder.end());

        preorder.erase(preorder.begin());
        vector<int> preR(preorder.begin()+i, preorder.end());
        preorder.erase(preorder.begin()+i, preorder.end());
        
        TreeNode* root = new TreeNode(curVal);
        root->left = buildTree(preorder, inorder);
        root->right = buildTree(preR, inR);
        return root;

    }
};