二叉搜索树的后续遍历

public class Solution {
    public boolean VerifySquenceOfBSTCore(int [] sequence, int start, int end){
        if(end-start<=1) return true;//当只有一个元素或者为空时，肯定满足条件
        int root=sequence[end-1];
        int splitIndex=start;
        for(;splitIndex<end-1;splitIndex++)
            if(sequence[splitIndex]>root)//找左右子树的划分点
                break;
        for(int i=splitIndex;i<end-1;i++)
            if(sequence[i]<root) //再找到划分点的情况下，若右子树还有元素小于根结点，说明不满足二叉搜索树的条件，返回false
                return false;
        boolean left=true;
//递归判断左右子树是不是二叉搜索树
        left=VerifySquenceOfBSTCore(sequence,start,splitIndex);
        boolean right=true;
        right=VerifySquenceOfBSTCore(sequence,splitIndex,end-1);
        return left&&right;
    }
    public boolean VerifySquenceOfBST(int [] sequence) {
        if(sequence==null||sequence.length<=0) return false;
        return VerifySquenceOfBSTCore(sequence, 0,sequence.length);
    }

}