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 a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (preorder == null) return null;
        
        int preLeft = 0, preRight = preorder.length - 1, inLeft = 0, inRight = inorder.length -1;
        return buildTree(preorder, inorder, preLeft, preRight, inLeft, inRight);
    }
    
    private TreeNode buildTree(int[] preorder, int[] inorder, int preLeft, int preRight, int inLeft, int inRight) {
        if (preLeft > preRight) return null;
        
        TreeNode temp = new TreeNode(preorder[preLeft]);
        if (preLeft == preRight)
            return temp;
        
        int index = inLeft;
        while (index < inRight && inorder[index] != preorder[preLeft])
            index++;
            
        temp.left = buildTree(preorder, inorder, preLeft + 1, preLeft + index - inLeft, inLeft, index - 1);
        temp.right = buildTree(preorder, inorder, preLeft + index - inLeft + 1, preRight, index + 1, inRight);
        return temp;
    }
}