PAT 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:

925 30 42 16 20 20 35 -5 28

Sample Output:

2 + 4 = 6

题目大意

有一棵空的BST，然后给你一个大小为N的序列，插入到BST中。求最底层和次底层的结点数。

解题思路

用数组tree[][4]模拟BST，tree[i][0]为i结点的数据，tree[i][1]为左孩子下标，tree[i][2]为右孩子下标，tree[i][3]为当前i结点的深度。

深搜过程模拟BST的插入，计算深度。

坑点：1、n=1时，应该输出1 + 0 = 1

#include <cstdio>#include <vector>#include <map>#include <set>#include <cstdlib>#include <climits>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define ll long longconst int MAXN = 1000 + 5;const int MAXM = 100000 + 5;const int INF = 0x7f7f7f7f;template <class XSD> inline XSD f_min(XSD a, XSD b) { if (a > b) a = b; return a; }template <class XSD> inline XSD f_max(XSD a, XSD b) { if (a < b) a = b; return a; }int n;int tree[MAXN][4];///0：数据 1：左子树 2：右子树 3：深度int Tdeep;void Dfs(int pos, int x, int xi, int deep){    if(tree[pos][0]>=x){///左子树        if(tree[pos][1]==-1){///为空            tree[pos][1] = xi;            tree[xi][3] = deep+1;            Tdeep = f_max(Tdeep, (deep+1));        }        else Dfs(tree[pos][1], x, xi, deep+1);///不为空    }else{///右子树        if(tree[pos][2]==-1){///为空            tree[pos][2] = xi;            tree[xi][3] = deep+1;            Tdeep = f_max(Tdeep, (deep+1));        }        else Dfs(tree[pos][2], x, xi, deep+1);///不为空    }}void Getdata(){    Tdeep = 0;    memset(tree, -1, sizeof tree);    scanf("%d", &tree[1][0]);    tree[1][3] = 1;    for(int i=2; i<=n; i++){        scanf("%d", &tree[i][0]);        Dfs(1, tree[i][0], i, 1);    }}void Solve(){    if(n==1){printf("1 + 0 = 1\n");return;}    int x=0, y=0;    for(int i=1; i<=n; i++){        if(tree[i][3]==Tdeep) x++;        else if(tree[i][3]==Tdeep-1) y++;    }    printf("%d + %d = %d\n", x, y, x+y);}int main(){    while(~scanf("%d", &n)){        Getdata();        Solve();    }    return 0;}