# PAT(Advanced Level)A1115. Counting Nodes in a BST

#### 思路

• 先按照BST来建树，建树之后用dfs来统计每一层的结点数，同时要用一个变量来记住最大深度，那么我们就可以得到最低两层的结点数量了

#### 代码

#include <iostream>
#include <vector>
#include <queue>
#include <map>
#include <string.h>
#include <set>
#include <unordered_map>
#include <algorithm>
using namespace std;
int cnt[1010] = {0};
int max_depth = -1;
struct node {
int val;
struct node* left;
struct node* right;
node(int x): val(x), left(NULL), right(NULL) {}
};
void insert(node*& root, int x) {
if(root == NULL) {
root = new node(x);
return;
}
if(x > root->val)
insert(root->right, x);
else
insert(root->left, x);
return;
}
void dfs(node* root, int depth) {
if(root) {
cnt[depth]++;
max_depth = max(max_depth, depth);
dfs(root->left, depth + 1);
dfs(root->right, depth + 1);
}
}
int main() {
int N, t;
cin >> N;
node* root = NULL;
while(N--) {
cin >> t;
insert(root, t);
}
dfs(root, 0);
int sum = cnt[max_depth - 1] + cnt[max_depth];
cout << cnt[max_depth] << " + " << cnt[max_depth - 1] << " = " << sum;
return 0;
}

posted @ 2021-03-14 16:38  MartinLwx  阅读(20)  评论(0编辑  收藏  举报