最后一个元素是 根节点。
左子树的元素都小于根节点，右子树都大于根节点
然后递归判断
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class Solution {
public:
bool VerifySquenceOfBST(vector<int> sequence) {
int size=sequence.size();
if(size==0)return false;
else if(size==1)return true;
int root=sequence.back();
int i,iLeftEnd;
bool result=true;
for(i=0;i<size-1;i++)
{
if(sequence[i]>root)
{
iLeftEnd=i;
break;
}
}
for(;i<size-1;i++)
{
if(sequence[i]<root)
return false;
}
if(iLeftEnd>0)//保证左子树有成员才判断，避免迭代空vector
{
vector<int>leftson;
leftson.assign(sequence.begin(),sequence.begin()+iLeftEnd);
result=VerifySquenceOfBST(leftson);
if(!result)return false;
}
if(iLeftEnd<size-1)//保证右子树有成员才判断，避免迭代空vector
{
vector<int>rightson;
rightson.assign(sequence.begin()+iLeftEnd,sequence.end()-1);
result=VerifySquenceOfBST(rightson);
}
return result;
}
}; |