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.

代码：

    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int len = preorder.size();
        if(len == 0)
            return NULL;
        TreeNode *root = new TreeNode(preorder[0]);
        int i;
        for(i = 0; i < len; i++){
            if(inorder[i] == preorder[0])
                break;
        }
        vector<int> lpreorder(preorder.begin()+1, preorder.begin()+i+1);
        vector<int> linorder(inorder.begin(), inorder.begin()+i);
        vector<int> rpreorder(preorder.begin()+i+1, preorder.end());
        vector<int> rinorder(inorder.begin()+i+1, inorder.end());
        vector<int>().swap(preorder);
        vector<int>().swap(inorder);
        root->left = buildTree(lpreorder, linorder);
        root->right = buildTree(rpreorder, rinorder);
        return root;
    }