PAT 甲级 1115 Counting Nodes in a BST

https://pintia.cn/problem-sets/994805342720868352/problems/994805355987451904

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

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

代码：

#include <bits/stdc++.h>
using namespace std;

int N;
vector<int> v(1000);
int depth = -1;

struct Node {
    int val;
    struct Node *left, *right;
};

Node* BuildBST(Node *root, int x) {
    if(!root) {
        root = new Node();
        root -> val = x;
        root -> left = NULL;
        root -> right = NULL;
    } else if(x <= root -> val)
        root -> left = BuildBST(root -> left, x);
    else root -> right = BuildBST(root -> right, x);

    return root;
}

void dfs(Node* root, int step) {
    if(!root) {
        depth = max(depth, step);
        return ;
    }
    v[step] ++;
    dfs(root -> left, step + 1);
    dfs(root -> right, step + 1);
}

int main() {
    scanf("%d", &N);
    Node *root = NULL;
    for(int i = 0; i < N; i ++) {
        int num;
        scanf("%d", &num);
        root = BuildBST(root, num);
    }
    dfs(root, 0);
    printf("%d + %d = %d
", v[depth - 1], v[depth - 2], v[depth - 1] + v[depth - 2]);
    return 0;
}

　　先建树然后 dfs 记录每一层的节点数目存在数组里和上个提交的题目比较类似了