leetcode--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.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        int length = inorder.length;
        return buildTreeHelp(preorder,inorder,0,length - 1, 0, length - 1);
    }
    
    public TreeNode buildTreeHelp(int[] preorder, int[] inorder, int pstart, int pend, int istart,int iend){
        TreeNode root = null;
        if(istart <= iend){
            root = new TreeNode(preorder[pstart]);
            int rootPosition = istart;
            int newpend = pstart;
            for(int i = istart; i <= iend; ++i){
                if(inorder[i] == preorder[pstart]){
                    rootPosition = i;
                    break;
                }
                else
                    ++newpend;                
            }
            root.left = buildTreeHelp(preorder, inorder, pstart + 1, newpend, istart, rootPosition - 1);
            root.right = buildTreeHelp(preorder, inorder, newpend + 1, pend, rootPosition + 1, iend);
        }
        return root;
    }
}