判断一个数组是否是一棵树的后续遍历

public boolean judge (int[] nums) {if (nums.length == 0 || nums == null)return true;int sum = 0;for (int i = 0; i < nums.length - 1; i++) {if (nums[i] >= nums[nums.length - 1]) {sum = i;break;}}int[] nums1 = new int[sum];for (int i = 0; i < nums1.length - 1; i++) {nums1[i] = nums[i];}int[] nums2 = new int[nums.length - 1 - sum];for (int i = nums1.length; i < nums.length -2; i++) {if (nums[i] < nums[nums.length - 1])return false;nums2[i - nums1.length] = nums[i];}boolean r1 = judge(nums1);boolean r2 = judge(nums2);if (r1 && r2)return true;return false;}

判断一个数组是否是一棵树的后续遍历，写代码写出感觉来了，停不下来。