二叉搜索树或者是一棵空树,或者是具有下列性质的二叉树: 若它的左子树不空,则左子树上所有结点的值均小于它的根结点的值;若它的右子树不空,则右子树上所有结点的值均大于它的根结点的值;它的左、右子树也分别为二叉搜索树。(摘自百度百科)
给定一系列互不相等的整数,将它们顺次插入一棵初始为空的二叉搜索树,然后对结果树的结构进行描述。你需要能判断给定的描述是否正确。例如将{ 2 4 1 3 0 }插入后,得到一棵二叉搜索树,则陈述句如“2是树的根”、“1和4是兄弟结点”、“3和0在同一层上”(指自顶向下的深度相同)、“2是4的双亲结点”、“3是4的左孩子”都是正确的;而“4是2的左孩子”、“1和3是兄弟结点”都是不正确的。
输入格式:
输入在第一行给出一个正整数N(≤100),随后一行给出N个互不相同的整数,数字间以空格分隔,要求将之顺次插入一棵初始为空的二叉搜索树。之后给出一个正整数M(≤100),随后M行,每行给出一句待判断的陈述句。陈述句有以下6种:
A is the root,即"A是树的根";
A and B are siblings,即"A和B是兄弟结点";
A is the parent of B,即"A是B的双亲结点";
A is the left child of B,即"A是B的左孩子";
A is the right child of B,即"A是B的右孩子";
A and B are on the same level,即"A和B在同一层上"。
题目保证所有给定的整数都在整型范围内。输出格式:
对每句陈述,如果正确则输出Yes,否则输出No,每句占一行。
输入样例:
5 2 4 1 3 0 8 2 is the root 1 and 4 are siblings 3 and 0 are on the same level 2 is the parent of 4 3 is the left child of 4 1 is the right child of 2 4 and 0 are on the same level 100 is the right child of 3
输出样例:
Yes Yes Yes Yes Yes No No No
#include <iostream>
#include <string>
#include <string.h>
#include <map>
using namespace std;
const int MAX = 1e7 + 10;
int tree[MAX]; //二叉搜索树
int deepth[MAX]; //结点深度
int tem;
map<int, int> num; //键值对应的结点编号
void creatTree(int x, int d) //建立二叉搜索树
{
if (tree[x] == 0x3f3f3f3f)
{
tree[x] = tem;
num.insert(make_pair(tem, x));
deepth[x] = d;
}
else
{
if (tem < tree[x])
creatTree(x * 2, d + 1);
else
creatTree(x * 2 + 1, d + 1);
}
return;
}
int main()
{
int n; cin >> n;
memset(tree, 0x3f, sizeof(tree)); //将每个元素初始化为0x3f3f3f3f
for (int i = 0; i < n; i++)
{
cin >> tem;
creatTree(1, 1);
}
int k; cin >> k;
while (k--)
{
string str;
int a, b;
cin >> a >> str;
if (str == "and")
{
cin >> b >> str >> str;
if (str == "siblings")
{
if (num.find(a) == num.end() || num.find(b) == num.end())
cout << "No" << endl;
else if (num[a] / 2 == num[b] / 2)
cout << "Yes" << endl;
else
cout << "No" << endl;
}
else
{
getline(cin, str);
if (num.find(a) == num.end() || num.find(b) == num.end())
cout << "No" << endl;
else if (deepth[num[a]] == deepth[num[b]])
cout << "Yes" << endl;
else
cout << "No" << endl;
}
}
else
{
cin >> str >> str;
if (str == "root")
{
if (num.find(a) == num.end())
cout << "No" << endl;
else if (num[a] == 1)
cout << "Yes" << endl;
else
cout << "No" << endl;
}
else if (str == "parent")
{
cin >> str >> b;
if (num.find(a) == num.end() || num.find(b) == num.end())
cout << "No" << endl;
else if (num[a] == num[b] / 2)
cout << "Yes" << endl;
else
cout << "No" << endl;
}
else if (str == "left")
{
cin >> str >> str >> b;
if (num.find(a) == num.end() || num.find(b) == num.end())
cout << "No" << endl;
else if (num[b] * 2 == num[a])
cout << "Yes" << endl;
else
cout << "No" << endl;
}
else
{
cin >> str >> str >> b;
if (num.find(a) == num.end() || num.find(b) == num.end())
cout << "No" << endl;
else if (num[b] * 2 + 1 == num[a])
cout << "Yes" << endl;
else
cout << "No" << endl;
}
}
}
return 0;
}
注意事项:
需要判断被询问的数据是否在树上,否则测试点2答案错误。
如有问题,欢迎提出。