剑指offer 二叉搜索树的后序遍历序列

 1 class Solution {
 2 public:
 3     bool VerifySquenceOfBST(vector<int> sequence) {
 4         int length = sequence.size();
 5         if( length == 0)
 6             return false;
 7         vector<int> less, greater;
 8         int temp = 0;
 9         //先将左子树压入less队列，并找到与右子树分割的结点temp
10         while(sequence[temp] < sequence.back())
11             less.push_back(sequence[temp ++]);
12         //在右子树中判断对于二叉搜索树的性质是否成立
13         while(temp < length){
14             if(sequence[temp] < sequence.back() )
15                 return false;
16             else
17                 greater.push_back(sequence[temp ++]);
18         }
19         //若子树为空，则说明遍历完毕，且性质符合
20         if(less.empty() || greater.empty())
21             return true;
22         return VerifySquenceOfBST(less) && VerifySquenceOfBST(greater);
23     }
24 };