# PAT Advanced 1115 Counting Nodes in a BST

## Overview

PAT Advanced 1115 Counting Nodes in a BST tests BST insertion and then BFS to get the number of nodes in the lowest 2 levels of the resulting tree.

## PAT Advenced 1115 Counting Nodes in a BST

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
```

## PAT Advenced 1115 Counting Nodes in a BST Solution

First construct the BST by insert each node in the order of the input, this takes O(N^2) time worst, but the input is <=1000 so, this should be fine, after geting the resulting tree, breadth first search the three and get the nodes of the lowest 2 levels. No tricky issues that we need to be concerned about.

The following is the C++ source code that could be accepted by PAT OJ to pass this PAT Advenced 1115 Counting Nodes in a BST:

## Summary

PAT Advanced 1115 Counting Nodes in a BST tests BST insertion and then BFS to get the number of nodes in the lowest 2 levels of the resulting tree.

Written on November 19, 2016