剑指offer 二叉搜索树后续遍历序列 判断

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

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39

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;                } }; 来源： http://tool.oschina.net/highlight