PAT (Advanced Level) Practice 1115 Counting Nodes in a BST (30 分) 凌宸1642

PAT (Advanced Level) Practice 1115 Counting Nodes in a BST (30 分) 凌宸1642

題目描述：

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.

譯：二叉搜索樹 (BST) 遞歸地定義爲具備如下屬性的二叉樹：

• 節點的左子樹僅包含鍵小於或等於節點鍵的節點。

• 節點的右子樹僅包含鍵大於節點鍵的節點。

• 左右子樹也必須是二叉搜索樹。

將一系列數字插入到最初爲空的二叉搜索樹中。 而後你應該計算結果樹的最低 2 個級別中的節點總數。

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Input Specification (輸入說明):

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.

譯：每一個輸入文件包含一個測試用例。 對於每種狀況，第一行給出一個正整數 N (≤1000)，它是輸入序列的大小。 而後在下一行給出 [−1000,1000] 中的 N 個整數，它們應該被插入到最初爲空的二叉搜索樹中。

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output Specification (輸出說明):

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.

譯：對於每種狀況，在一行中打印結果樹的最低 2 級中的節點數，格式以下：

n1 + n2 = n

其中 n1 是最低層的節點數，n2 是上層的節點數，n 是總和。

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Sample Input (樣例輸入):

9
25 30 42 16 20 20 35 -5 28

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Sample Output (樣例輸出):

2 + 4 = 6

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The Idea：

• 節點數很少，利用鏈式二叉樹結構，根據輸入的數據，簡歷二叉搜索樹 BST。
• 對二叉搜索樹進行層序遍歷，而且記錄層數以及每層節點的個數。
• 輸出最後兩層的算術和的算式。

The Codes:

#include<bits/stdc++.h>
using namespace std ;
const int maxn = 1010 ;
int n , t , cnt = 0 ;
vector<int> ans ;
struct node{
	int val ;
	node* l ;
	node* r ;
};
void insert(node* &root , int data){
	if(!root){
		root = new node() ;
		root->val = data ;
		root->l = root->r = NULL ;
		return ;   // 必定要記得返回，否則會爆棧 
	}
	if(data <= root->val) insert(root->l , data) ;	
	else insert(root->r , data) ;	
}
void bfs(node* root){
	queue<node*> q ;
	q.push(root) ;
	while(!q.empty()){
		cnt ++ ;  // 記錄總共有幾層 
		int size = q.size() ;
		ans.push_back(size );
		for(int i = 0 ; i < size ; i ++){
			node* top = q.front() ;
			q.pop() ;
			if(top->l != NULL) q.push(top->l) ;  	// 左孩子不空，則入隊 
			if(top->r != NULL) q.push(top->r) ;		// 右孩子不空，則入隊 
		}
		
	}
}
int main(){
	cin >> n ;	
	node* root = NULL ;
	for(int i = 0 ; i < n ; i ++){
		cin >> t ;
		insert(root , t) ;
	}
	bfs(root) ;
	printf("%d + %d = %d\n" , ans[cnt - 1] , ans[cnt - 2] , ans[cnt - 1] + ans[cnt - 2]) ;
	return 0 ;
}