二叉搜索树的后序遍历序列:递归中不要过早return f(左)&&f(右)

题目描述

class Solution {
public:
bool VerifySquenceOfBST(vector<int> sequence) {
if(sequence.empty()) return false;
return isSquenceOfBST(sequence,0,sequence.size()-1);
}
bool isSquenceOfBST(vector<int>&sequence,int start,int end){

int i=start;
for (; i < end; ++i)
{
if(sequence[i]>sequence[end]) break;
}
int j=i;
for (; j < end; ++j)
{
if(sequence[j]<sequence[end]) return false;
}
bool left=true;
if(i>start) //举例，如果是[10,7,5]，左子树为空，所以左边返回true
left=isSquenceOfBST(sequence,start,i-1);
bool right=true;
if(i<end) //举例，如果是[1,2,3,4]，右子树为空，所以右边返回true
right=isSquenceOfBST(sequence,i,end-1);

return left&&right;

}
};