LeetCode - 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.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if(inorder == null || inorder.length == 0) {
            return null;
        }
        TreeNode root = new TreeNode(postorder[postorder.length-1]);
        int mid = 0;
        for (int i = 0; i < inorder.length; i++) {
            if (inorder[i] == postorder[postorder.length-1]) {
                mid = i;
                break;
            }
        }
        root.left = buildTree(Arrays.copyOfRange(inorder, 0, mid), Arrays.copyOfRange(postorder, 0, mid));
        root.right = buildTree(Arrays.copyOfRange(inorder, mid+1, inorder.length), Arrays.copyOfRange(postorder, mid, postorder.length-1));
        return root;
    }
}