PAT Advanced 1115 Counting Nodes in a BST (30) [⼆叉树的遍历，BFS，DFS]

题目

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: The lef 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 lef 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

题目分析

已知一棵二叉查找树的建树序列，求二叉查找树最后两层的结点数

解题思路

思路 01

1. 建树（非递归）
2. bfs深度优先遍历，记录每层结点数和最大层数

思路 02

1. 建树（递归）
2. dfs广度优先遍历，记录每层结点数和最大层数

Code

Code 01

#include <iostream>
#include <queue>
using namespace std;
const int maxn=1000;
struct node {
	int data;
	int h=0;
	node * left=NULL;
	node * right=NULL;
	node() {
	}
	node(int _data, int _h) {
		data = _data;
		h=_h;
	}
};
node * root;
int h[maxn],lcn[maxn],max_h;
void insert(int n) {
	if(root==NULL) {
		root=new node(n,1);
		return;
	}
	node * p=root;
	while(p!=NULL) {
		if(p->data<n) {
			if(p->right==NULL){
				p->right=new node(n,p->h+1);
				return;
			}
			else p=p->right;
		} else {
			if(p->left==NULL){
				p->left=new node(n,p->h+1);
				return;
			}
			else p=p->left;
		}
	}
}
void bfs() {
	queue <node *> q;
	q.push(root);
	while(!q.empty()) {
		node * now = q.front();
		q.pop();
		max_h=max(max_h,now->h);
		lcn[now->h]++;
		if(now->left!=NULL)q.push(now->left);
		if(now->right!=NULL)q.push(now->right);
	}
}
int main(int argc,char * argv[]) {
	int n,rn;
	scanf("%d",&n);
	for(int i=0; i<n; i++) {
		scanf("%d",&rn);
		insert(rn);
	}
	bfs();
	//根节点高度为1，高度0作为哨兵
	printf("%d + %d = %d", lcn[max_h], lcn[max_h-1], lcn[max_h]+lcn[max_h-1]);
	return 0;
}

Code 02

#include <iostream>
#include <queue>
using namespace std;
const int maxn=1000;
struct node {
	int data;
	int h=0;
	node * left=NULL;
	node * right=NULL;
	node() {
	}
	node(int _data, int _h) {
		data = _data;
		h=_h;
	}
};
node * root;
int h[maxn],lcn[maxn],max_h;
void insert(node* &root, int data, int dep) {
    if(root == NULL) {      //到达空结点时，即为需要插入的位置
        root = new node(data,dep);
        root->data = data;
        root->left = root->right = NULL;    //此句不能漏
        return;
    }
    if(data <= root->data) insert(root->left, data, root->h+1);     //插在左子树
    else insert(root->right, data, root->h+1);     //插在右子树
}
void dfs(node * now, int h){
	if(now==NULL){
		return;
	}
	max_h=max(max_h,h);
	lcn[h]++;
	if(now->left!=NULL)dfs(now->left, h+1);
	if(now->right!=NULL)dfs(now->right, h+1);
}
int main(int argc,char * argv[]) {
	int n,rn;
	scanf("%d",&n);
	for(int i=0; i<n; i++) {
		scanf("%d",&rn);
		insert(root,rn, 1);
	}
	dfs(root,1);
	//根节点高度为1，高度0作为哨兵
	printf("%d + %d = %d", lcn[max_h], lcn[max_h-1], lcn[max_h]+lcn[max_h-1]);
	return 0;
}