PAT (Advanced Level) Practice 1115 Counting Nodes in a BST (30分) （不用BFS、DFS、不用遍历直接建树标记完事）

1.题目

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 [−10001000] 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

2.题目分析

建树的时候加上层数概念，题目中求的是二叉搜索树的最下面一行以及这一行的上一行，我设根节点深度为0，即输出深度最大以及第二大的节点个数就行（使用map标记，无需遍历）

3.代码

#include<iostream>
#include<map>
using namespace std;
typedef struct node * tree;
struct node
{
	int data;
	tree left;
	tree right;
};
map<int,int>levels;
tree creat(tree t, int x,int level)
{
	if (!t)
	{
		t = (tree)malloc(sizeof(struct node));
		t->data = x;
		t->left = t->right = NULL;
		levels[level]++;
		return t;
	}
	if (t->data < x)
		t->right = creat(t->right, x, level + 1);
	else
		t->left = creat(t->left, x, level + 1);
	return t;
}

int main()
{
	int n,a;
	scanf("%d", &n);
	tree t = NULL;
	for (int i = 0; i < n; i++)
	{
		scanf("%d", &a);
		t = creat(t, a, 0);
	}
	int size = levels.size();
	printf("%d + %d = %d\n", levels[size - 1], levels[size - 2], levels[size - 1] + levels[size - 2]);

}