PAT (Advanced Level)1115. Counting Nodes in a BST (30)

1115. Counting Nodes in a BST (30)

time limit

400 ms

memory limit

65536 kB

code length limit

16000 B

Judgment procedure

Standard

author

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

//1115. Counting Nodes in a BST(30)

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int a[1001] = { 0 }, level=-1;
typedef struct node {
	int data;
	node *l, *r;
}*Tree;
Tree T;

void createtree(Tree &T,int num) {
	if (T == NULL) {
		T = new node();
		T->data = num;
		T->l = NULL;
		T->r = NULL;
		return;
	}
	if (num <= T->data) 
		createtree(T -> l, num);
	else createtree(T->r, num);
}

void dfs(Tree T,int depth) {
	if (T == NULL) return;
	a[depth]++;
	level = max(level, depth);
	dfs(T->l, depth+1);
	dfs(T->r, depth+1);
}

int main() {
	int n,in[1001];
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> in[i];
		createtree(T, in[i]);
	}
	dfs(T, 0);
	printf("%d + %d = %d\n", a[level], a[level - 1], a[level] + a[level - 1]);
	return 0;
}