1115 Counting Nodes in a BST （30 分）

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 [−10001000] 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

題意：

給定n個數，建一棵二叉搜尋樹。問倒數第一層和倒數第二層分別有多少個節點。

思路：

先建樹，然後dfs看每個節點的深度。

 1 #include <iostream>
 2 #include <set>
 3 #include <cmath>
 4 #include <stdio.h>
 5 #include <cstring>
 6 #include <algorithm>
 7 #include <vector>
 8 #include <queue>
 9 #include <map>
10 #include <bits/stdc++.h>
11 using namespace std;
12 typedef long long LL;
13 #define inf 0x7f7f7f7f
14 
15 const int maxn = 1005;
16 int n, num[maxn], cnt[maxn], dep = -1;
17 struct node{
18     int val;
19     int left = -1, right = -1;
20     int height;
21 }tree[maxn];
22 int tot;
23 
24 void add(node t, int rt)
25 {
26     if(t.val > tree[rt].val && tree[rt].right != -1){
27         add(t, tree[rt].right);
28     }
29     else if(t.val > tree[rt].val){
30         tree[tot] = t;
31         tree[rt].right = tot++;
32     }
33     else if(tree[rt].left != -1){
34         add(t, tree[rt].left);
35     }
36     else{
37         tree[tot] = t;
38         tree[rt].left = tot++;
39     }
40 }
41 
42 void dfs(int rt, int h)
43 {
44     tree[rt].height = h;
45     cnt[h]++;
46     dep = max(dep, h);
47     if(tree[rt].left != -1){
48         dfs(tree[rt].left, h + 1);
49     }
50     if(tree[rt].right != -1){
51         dfs(tree[rt].right, h + 1);
52     }
53 }
54 
55 int main()
56 {
57     scanf("%d", &n);
58     scanf("%d", &tree[tot++].val);
59     for(int i = 1; i < n; i++){
60         node t;
61         scanf("%d", &t.val);
62         add(t, 0);
63     }
64 
65     //printf("!\n");
66     dfs(0, 1);
67     int n1 = cnt[dep], n2 = cnt[dep - 1];
68     printf("%d + %d = %d\n", n1, n2, n1 + n2);
69     return 0;
70 }