1115 Counting Nodes in a BST (30 point(s))

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本文介绍了一种算法，用于根据给定序列生成二叉搜索树(BST)，并计算树最低两层的节点数量。通过递归插入和深度优先搜索(DFS)遍历，实现了对BST的构建与节点计数。

1115 Counting Nodes in a BST (30 point(s))

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 [−10001000] 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

中规中矩的压轴题。非常开心，25分钟AC本题。

题意：

根据给定的序列生成二叉搜索树，并计算最后两层的结点数。

考察点：

二叉搜索树的建立、树的遍历

思路：

1. 定义结点结构，建立BST；

2. 采用DFS遍历整棵树，搜索过程中把所在层次level作为参数，注意对于每个结点记录其所在层次。

注意点：

1. Insert()参数root必须是结点指针的引用，否则建立动态建立新结点的时候会发生错误。

2. 注意这种记录每个层次结点树的方法。

2. 1099 https://blog.youkuaiyun.com/coderwait/article/details/100183241

#include<iostream>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
struct Node{
	int data;Node* left;Node* right;
	Node(int d):data(d),left(NULL),right(NULL){	}
};
int levelCnt[1007]={0};
void Insert(Node* &root,int data){
	if(root==NULL){
		root = new Node(data);
		return;
	}
	if(data<=root->data) Insert(root->left,data);
	else Insert(root->right,data);
}
int maxLevel=0;
void dfs(Node* &root,int level){
	if(root!=NULL){
		levelCnt[level]++;
		maxLevel = max(maxLevel,level);
	}
	if(root->left!=NULL){
		dfs(root->left,level+1);
	}
	if(root->right!=NULL){
		dfs(root->right,level+1);
	}
}
int main(void){
	int N,a;cin>>N;
	Node* root = NULL;
	for(int i=0;i<N;i++){
		cin>>a;
		Insert(root,a);
	}
	dfs(root,0);
	cout<<levelCnt[maxLevel]<<" + "<<levelCnt[maxLevel-1]<<" = "<<levelCnt[maxLevel]+levelCnt[maxLevel-1]<<endl;
	return 0;
}