算法练习---从中序与后序遍历序列构造二叉树

题目:

给定两个整数数组inorder和postorder ，其中inorder是二叉树的中序遍历，postorder是同一棵树的后序遍历，请构造并返回这颗二叉树
示例:
输入：inorder = [9, 3, 15, 20, 7], postorder = [9, 15, 7, 20, 3]
输出：[3, 9, 20, null, null, 15, 7]
示例:
输入：inorder = [-1], postorder = [-1]
输出：[-1]
代码:

 public TreeNode<Integer> buildTree(int[] inorder, int[] postorder) {
        if (inorder == null || postorder == null || inorder.length == 0 || postorder.length == 0) {
            return null;
        }
        TreeNode<Integer> node = new TreeNode<>(postorder[postorder.length - 1]);
        int index = -1;
        int index2 = -1;
        for (int x = 0; x < inorder.length; x++) {
            //后序特点是 根在最后,中序特点是根左边是左子树,根右边是右子树
            //获取后序数组中最后一位值,在中序数组中找出值的位置,并记录index
            //获取在中序中index+1在的值,找出这个值在后序中的位置并记录 index2
            //最后切割两个数组.
            if (postorder[postorder.length - 1] == inorder[x]) {
                index = x;
                int[] leftI = null;
                int[] rightI = null;
                int[] leftP = null;
                int[] rightP = null;
                for (int y = 0; y < postorder.length; y++) {
                    int temp = -1;
                    if (index + 1 < inorder.length) {
                        temp = index + 1;
                    } else if (index + 1 == inorder.length) {
                        temp = index;
                    }
                    if (inorder[temp] == postorder[y]) {
                        index2 = y;
                        leftI = Arrays.copyOfRange(inorder, 0, index);
                        rightI = Arrays.copyOfRange(inorder, index + 1, inorder.length);
                        leftP = Arrays.copyOfRange(postorder, 0, index2);
                        rightP = Arrays.copyOfRange(postorder, index2, postorder.length - 1);
                        break;
                    }
                }
                node.lTreeNode = buildTree(leftI, leftP);
                node.rTreeNode = buildTree(rightI, rightP);
            }
        }
        return node;
    }