给出一颗BST树 求这个数最后两层的节点数量

用c1+c2 = n 的形式给出

建树深度搜索即可

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
using namespace std;struct node{int l,r,x;
}n[2010];
int tag,root=1,step,c1,c2;int insert(int rt,int val,int st){step = max(step,st);if(rt!=1111){if(val<=n[rt].x)n[rt].l = insert(n[rt].l,val,st+1);else n[rt].r = insert(n[rt].r,val,st+1);return rt;//******if empty here!}else{tag++;n[tag].x = val;n[tag].l = n[tag].r = 1111;return tag;}
}
void dfs(int rt,int st){if(rt!=1111){if(n[rt].l!=1111)dfs(n[rt].l,st+1);if(st==step)c1++;else if(st==step-1)c2++;if(n[rt].r!=1111)dfs(n[rt].r,st+1);}
}
int main()
{int N;cin>>N;root = 1111;for(int i=1;i<=N;i++){int val;cin>>val;root = insert(root,val,1);}dfs(root,1);cout<<c1<<" + "<<c2<<" = "<<c1+c2<<endl;return 0;
}