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:二叉搜尋樹 (BST) 遞迴地定義為具有以下屬性的二叉樹:
-
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尋樹。
將一系列數字插入到最初為空的二叉搜尋樹中。 然後你應該計算結果樹的最低 2 個級別中的節點總數。
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:每個輸入檔案包含一個測試用例。 對於每種情況,第一行給出一個正整數 N (≤1000),它是輸入序列的大小。 然後在下一行給出 [−1000,1000] 中的 N 個整數,它們應該被插入到最初為空的二叉搜尋樹中。
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:對於每種情況,在一行中列印結果樹的最低 2 級中的節點數,格式如下:
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。
Sample Input (樣例輸入):
9
25 30 42 16 20 20 35 -5 28
Sample Output (樣例輸出):
2 + 4 = 6
The Idea:
- 節點數不多,利用鏈式二叉樹結構,根據輸入的資料,簡歷二叉搜尋樹 BST。
- 對二叉搜尋樹進行層序遍歷,並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 ; // 一定要記得返回,不然會爆棧
}
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 ;
}