1115. Counting Nodes in a BST (30)-PAT甲级真题（二叉树的遍历，dfs）

1115. Counting Nodes in a BST (30)
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：输出一个二叉搜索树的最后两层结点个数a和b，以及他们的和c：“a + b = c”

分析：用链表存储，递归构建二叉搜索树，深度优先搜索，传入的参数为结点和当前结点的深度depth，如果当前结点为NULL就更新[b]最大深度maxdepth的值并return，将每一层所对应的结点个数存储在数组num中，输出数组的最后两个的值~~~~[/b]

#include <cstdio>
#include <vector>
using namespace std;
struct node {
int v;
struct node *left, *right;
};
node* build(node *root, int v) {
if(root == NULL) {
root = new node();
root->v = v;
root->left = root->right = NULL;
} else if(v <= root->v)
root->left = build(root->left, v);
else
root->right = build(root->right, v);
return root;
}
vector<int> num(1000);
int maxdepth = -1;
void dfs(node *root, int depth) {
if(root == NULL) {
maxdepth = max(depth, maxdepth);
return ;
}
num[depth]++;
dfs(root->left, depth + 1);
dfs(root->right, depth + 1);

}
int main() {
int n, t;
scanf("%d", &n);
node *root = NULL;
for(int i = 0; i < n; i++) {
scanf("%d", &t);
root = build(root, t);
}
dfs(root, 0);
printf("%d + %d = %d", num[maxdepth-1], num[maxdepth-2], num[maxdepth-1] + num[maxdepth-2]);
return 0;
}