Construct Binary Tree from Preorder and Inorder Traversal

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: Recursion.

/**
 * 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) {
        return buildTreeRe(preorder.begin(), inorder.begin(), preorder.size());
    }
    
    TreeNode* buildTreeRe(vector<int>::iterator preorder, vector<int>::iterator inorder, int N) 
    {
        if(N <= 0) return NULL;
        vector<int>::iterator it = find(inorder, inorder + N, *preorder);
        int pos = it - inorder;
        TreeNode* root = new TreeNode(*preorder);
        root->left = buildTreeRe(preorder+1, inorder, pos);
        root->right = buildTreeRe(preorder+pos+1, inorder+pos+1, N-1-pos);
        return root;
    }
};