1115 Counting Nodes in a BST (30 分)

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 28

Sample Output:

2 + 4 = 6

题目大意：给出排序二叉树序列，构造出二叉树后，求出最底下两层的数目和；
思路：链表存储，建树时统计最低两层的结点数

#include<iostream>
using namespace std;
struct node {
	int data, layer;
	node* lchild, * rchild;
};
int lowest = -1, cnt1 = 0, above, cnt2 = 0;
node* create(node*& root, int data, int layer) {
	if (root == NULL) {
		root = new node;
		root->data = data;
		root->lchild = root->rchild = NULL;
		if (layer > lowest) {
			lowest = layer;
			above = layer - 1;
			cnt2 = cnt1;
			cnt1 = 1;
		}
		else if (layer == lowest) {
			++cnt1;
		}
		else if (layer == above) {
			++cnt2;
		}
	}
	else if (data <= root->data) root->lchild = create(root->lchild, data, layer + 1);
	else root->rchild = create(root->rchild, data, layer + 1);
	return root;
}
int main() {
	int n;
	cin >> n;
	node* root = NULL;
	for (int i = 0;i < n;++i) {
		int data;
		cin >> data;
		root = create(root, data, 0);
	}
	cout << cnt1 << " + " << cnt2 << " = " << cnt1 + cnt2 << endl;
	return 0;
}