浙大2016上机考研题

D. Counting Nodes in a BST (30)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CHEN, Yue

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

无奈被吞一次；

题意：建立bst树，用bfs访问最下2层节点数；

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <queue>
using namespace std;
struct bst{
int data;
bst *left;
bst *right;
}*bbshu;
int level[1000]={0};
int height=0;

bst* inser(int k,bst *p)
{
if(!p)
{
p=new bst;
p->data=k;
p->right=p->left=NULL;
}else if(p->data>=k)
{
p->left=inser(k,p->left);
}
else if(p->data<k)
{
p->right=inser(k,p->right);
}
return p;
}
//树的层序遍历，就是bfs，用指针写的
void bfs(bst *p)
{
queue<bst*> Q;
if(!p) return;
Q.push(p);
height=0;
bst* last=p;
bst* tail=p;
level[height]=1;
height=1;
while(!Q.empty())
{
bst* temp=Q.front();
if(temp==NULL) break;
int k=temp->data;
if(temp->left!=NULL)
{
Q.push(temp->left);
level[height]++;
}
if(temp->right!=NULL)
{
Q.push(temp->right);
level[height]++;
}
Q.pop();
if(k==tail->data)
{
tail=Q.back();
last=tail;
height++;
}
}
}
int main()
{
int k;
cin>>k;
bbshu=NULL;
bst *point=bbshu;
for(int i=0;i<k;i++)
{
int num;
cin>>num;
point=inser(num,point);
}

bfs(point);
int k1=level[height-2];
int k2=level[height-3];
cout<<k1<<" + "<<k2<<" = "&l
9b6c
t;<k1+k2<<endl;
return 0;
}