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PAT甲级1115

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1115. Counting Nodes in a BST (30) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 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 proper


1115. Counting Nodes in a BST (30)


时间限制



400 ms



内存限制



65536 kB



代码长度限制



16000 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



#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
struct Node
{
int data;
int depth;
Node *l, *r;
Node(int x) :data(x), l(NULL), r(NULL) {}
}*root;
void insert(Node*&root, int x)//这个地方要用引用
{
if (root == NULL)
{
root = new Node(x);
return;
}
else if (x <= root->data)
{
insert(root->l, x);
}
else
insert(root->r, x);
}
int maxdepth = -1;
void dfs(Node*root, int depth)
{
if (!root)
{
maxdepth = max(maxdepth, depth);
return;
}
dfs(root->l, depth + 1);
dfs(root->r, depth + 1);
}
int sum1=0, sum2=0;
void levelorder(Node*root)
{
queue<Node*> Q;
Q.push(root);
root->depth = 1;
if (root->depth == maxdepth)
{
sum2++;
}
else if (root->depth == maxdepth - 1)
sum1++;
while (!Q.empty())
{
Node*f = Q.front();
Q.pop();
if (f->l)
{
Q.push(f->l);
f->l->depth = f->depth + 1;
if (f->l->depth == maxdepth - 1)
{
sum1++;
}
else if (f->l->depth == maxdepth)
{
sum2++;
}
}
if (f->r)
{
Q.push(f->r);
f->r->depth = f->depth + 1;
if (f->r->depth == maxdepth - 1)
{
sum1++;
}
else if (f->r->depth == maxdepth)
{
sum2++;
}
}
}
}
int main()
{
int n;
scanf("%d", &n);
int x;
for (int i = 0; i < n; i++)
{
scanf("%d", &x);
insert(root, x);
}
dfs(root, 0);
levelorder(root);
printf("%d + %d = %d\n", sum2, sum1, sum1 + sum2);
return 0;
}



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