判断是否是二叉搜索树的后序遍历

输入一个整数数组，判断该数组是不是某二叉搜索树的后序遍历的结果。假设输入的数组的任意两个数字都互不相同。

5 7 6 9 11 10 8  

Yes

7 4 6 5

No

public class Solution {
    
    public boolean VerifySquenceOfBST(int [] sequence) {

        if(sequence.length==0) return false;
        return F(sequence, 0, sequence.length-1);

    }


    public boolean F(int[] sequence, int start, int end){

        if(start>= end) return true;

        //从后往前找到左右子树分割的位置
        int rightIndex = end-1;
        while(rightIndex>=start){
            if(sequence[rightIndex]>sequence[end]) rightIndex--;
            else break;
        }
        rightIndex++;
        //继续往前，如果符合二叉搜索树，应该是每一个都比自己小
        int leftIndex = rightIndex-1;
        while(leftIndex>=start){
            if(sequence[leftIndex]<sequence[end]) leftIndex--;
            else break;
        }

        //如果leftIndex!=start-1说明左子树有比自己大的
        if(leftIndex != start-1){
            return false;
        }
        else{
            return F(sequence, start, rightIndex-1) && F(sequence, rightIndex, end-1);
        }
    }
}