Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree. For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        return treeHelper(begin(postorder), end(postorder), begin(inorder), end(inorder));
    }

private:
    // vector<int>::iterator ps_first, ps_end, in_first, in_end;
    template <typename IT>
    TreeNode* treeHelper(IT ps_first, IT ps_end, IT in_first, IT in_end) {
       if(ps_first == ps_end || in_first == in_end) return nullptr;
        int rtval = *prev(ps_end);
        TreeNode* root = new TreeNode(rtval);

        //vector<int>::iterator index;
        auto index = find(in_first, in_end, rtval);
        // left subtree size
        int lsize = distance(in_first, index);

        root->left = treeHelper(ps_first, next(ps_first, lsize),
                                in_first, index);
        // bug 1, index is the iterator of vector<int> inorder, not post order
        // bug 2, prev(ps_end) point to the last element, not ps_end(point to the element after the last element)
        //root->right = treeHelper(next(ps_first, lsize), index,
        root->right = treeHelper(next(ps_first, lsize), prev(ps_end),
                                 next(index), in_end);

        return root;
    }
};