原题链接：1115 Counting Nodes in a BST (30分)
关键词：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
题目大意： 给出一组数构建BST，让你输出最下面两层的结点数。
分析： 建树，然后用dfs遍历，递归到孩子就将h+1，并用数组记录每个高度的节点数。
代码：
#include <bits/stdc++.h>
using namespace std;const int maxn = 1e4 + 10;struct node{int val, h;node *l, *r;
};node* insert(node* root, int value){ //插入建树if(root == NULL){root = new(node);root->val = value;root->l = root->r = NULL;}else{if(value <= root->val) root->l = insert(root->l, value);else root->r = insert(root->r, value);}return root;
}int mx = -1; //mx-1为最大高度
int num[maxn]; //记录每个高度的节点数
void dfs(node* root, int h){if(root == NULL){mx = max(mx, h);return;}num[h]++;dfs(root->l, h + 1);dfs(root->r, h + 1);
}int main(){int n;scanf("%d", &n);node* root = NULL;for(int i = 0; i < n; i ++ ){int x;scanf("%d", &x);root = insert(root, x);}dfs(root, 1);//printf("%d\n", mx);printf("%d + %d = %d", num[mx-1], num[mx-2], num[mx-1] + num[mx-2]);return 0;
}