题目描述：
A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:
- The left subtree of a node contains only nodes with keys less than or equal to the node's key.
- The right subtree of a node contains only nodes with keys greater than the node's key.
- Both the left and right subtrees must also be binary search trees.
Insert a sequence of numbers into an initially empty binary search tree. Then you are supposed to count the total number of nodes in the lowest 2 levels of the resulting tree.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (<=1000) which is the size of the input sequence. Then given in the next line are the N integers in [-1000 1000] which are supposed to be inserted into an initially empty binary search tree.
Output Specification:
For each case, print in one line the numbers of nodes in the lowest 2 levels of the resulting tree in the format:
n1 + n2 = n
where n1 is the number of nodes in the lowest level, n2 is that of the level above, and n is the sum.
Sample Input:9 25 30 42 16 20 20 35 -5 28Sample Output:
2 + 4 = 6
题目思路：
首先建BST树，题目要求输出最后一层和倒数第二层的结点的个数。我们可以用层次遍历的方法，直接记录最后2层的个数。或者直接深搜，记录每个深度的结点个数。这里给出2种方法的代码。
题目代码：
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
int ans[10000], maxd;
// Bst结构
struct Bst{int v;Bst *lchild, *rchild;Bst(int v):v(v),lchild(NULL),rchild(NULL){}
};
// 插入建树
void insert(Bst *&root, int v){if(root == NULL){root = new Bst(v);return ;}if(v > root->v){insert(root->rchild, v);}else{insert(root->lchild, v);}
}void dfs(Bst *root, int d){if(root == NULL ) return ;ans[d]++; // 记录每个深度的结点数 maxd = max(maxd,d);dfs(root->lchild,d+1);dfs(root->rchild,d+1);
}int n, v, l1 ,l2;
int main(){scanf("%d",&n);memset(ans,0,sizeof(ans));Bst *root = NULL;while(n--){scanf("%d",&v);insert(root,v);}// 深搜 maxd = -1;dfs(root, 0);queue<Bst*>q;q.push(root);l1 = l2 = 0;// 层次遍历方法 l1 l2表示最底层和倒数第二层个数 while(!q.empty()){l2 = l1;l1 = q.size();for(int i = 0; i < l1; i++){ //注意这里 Bst *p = q.front();if(p->lchild != NULL) q.push(p->lchild);if(p->rchild != NULL) q.push(p->rchild); q.pop();}} printf("%d + %d = %d\n",ans[maxd],ans[maxd-1],ans[maxd]+ans[maxd-1]);
// printf("%d + %d = %d\n",l1,l2,l1+l2);return 0;
}