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

思路

简单题,打印出二叉搜索树最后两层的节点数之和,格式为:"n1 + n2 = sum(n1,n2)"。

1.根据输入建立二叉搜索树
2.前序遍历二叉树,用一个数组统计每一层的节点数,levels[i]代表第i层的节点数,用一个maxlevel不断更新最大层数。
3.输出levels[maxlevel] + levels[maxlevel - 1]。 代码
#include<iostream>
#include<vector>
using namespace std;
vector<int> levels;
int maxlevel = -1;
class node
{
public:
node(int _val):val(_val),left(nullptr),right(nullptr)
{
}
int val;
node* left;
node* right;
}; typedef node* treenode; void buildBST(treenode& root,int val)
{
if(val <= root->val && !root->left)
{
root->left = new node(val);
return;
}
if(val <= root->val && root->left)
{
buildBST(root->left,val);
return;
}
if(val > root->val && !root->right)
{
root->right = new node(val);
return;
}
if(val > root->val && root->right)
{
buildBST(root->right,val);
return;
}
} void countNodes(const treenode root,int level)
{
levels[level]++;
if(level > maxlevel)
maxlevel = level;
if(!root->left && !root->right)
return;
if(root->left)
countNodes(root->left,level + 1);
if(root->right)
countNodes(root->right,level + 1);
} int main()
{
int n;
while(cin >> n)
{
levels.resize(n + 1);
treenode root = new node(0);
cin >> root->val;
for(int i = 1;i < n;i++)
{
int tmp;
cin >> tmp;
buildBST(root,tmp);
}
countNodes(root,1);
cout << levels[maxlevel] << " + " << levels[maxlevel - 1] << " = " << levels[maxlevel] + levels[maxlevel - 1] << endl;
}
}

  

 

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