剑指offer 24：二叉搜索树的后序遍历序列

class Solution {
public:
    bool VerifySquenceOfBST(vector<int> sequence) {
        int size = sequence.size();
        if(size ==0){
            return false;
        }
        return VerifyPostOfBST( sequence,0,size-1);
    }
    bool VerifyPostOfBST(vector<int> sequence,int start,int end){
        if(start==end){
            return true;
        }
        int root=sequence[end];
        int mid=end;
        for(int i=start ; i<end ; i++){
            if(sequence[i]>=sequence[end]){
                mid=i;
                break;
            }
        }
        bool curr=true;
        //根节点没有左子树或右子树的情况
        bool v1=true,v2=true;
        if(mid==start){
            for(int i= mid; i<end;i++){
                 curr = curr && sequence[i]>=sequence[end];
            }
            v2=VerifyPostOfBST(sequence,mid,end-1);
        }else if(mid==end){
            for(int i= start; i<mid;i++){
                curr = curr && sequence[i]<=sequence[end];
            }
            v1=VerifyPostOfBST(sequence,start,mid-1);
        }else {
            for(int i= mid; i<end;i++){
                 curr = curr && sequence[i]>=sequence[end];
            }
            for(int i= start; i<mid;i++){
                curr = curr && sequence[i]<=sequence[end];
            }
            v1=VerifyPostOfBST(sequence,start,mid-1);
            v2=VerifyPostOfBST(sequence,mid,end-1);
        }
        
        return v1 && v2 && curr;
    }
};