题目描述

输入样例1
6
8 27 5 1
9 40 -1 -1
10 20 0 3
12 21 -1 4
15 22 -1 -1
5 35 -1 -1
输出样例1
YES
输入样例2
6
8 27 5 1
9 40 -1 -1
10 20 0 3
12 11 -1 4
15 22 -1 -1
50 35 -1 -1
输出样例2
NO

思路

见注释

AC代码

#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
typedef struct node
{int data;struct node *left, *right;
}node;
int cnt; //记录是否能建立起一棵树
int num[N][2]; //存储每个节点的左子树和右子树的键值
int k1[N], k2[N]; //存储每个节点的K1和K2键值
bool flag_k2, flag_k1; //比较每个节点的左右子树是否都满足情况
vector<int> mid;
node* Init(int pos)
{if(pos == -1) return NULL; //当前节点不存在cnt ++;node* temp = new node;temp->data = pos;temp->left = NULL;temp->right = NULL;temp->left = Init(num[pos][0]); //左子树键值temp->right = Init(num[pos][1]); //右子树键值return temp;
}void midorder(node* tree)
{if(tree){midorder(tree->left);mid.push_back(tree->data);midorder(tree->right);}
}void Judge(node* tree)
{//小根堆的特点，该节点左右子树的值均>该节点，即根节点的值最小if(tree->left){if(k2[tree->data] > k2[tree->left->data]){flag_k2 = false;return;}Judge(tree->left);}if(tree->right){if(k2[tree->data] > k2[tree->right->data]){flag_k2 = false;return;}Judge(tree->right);}
}
int main()
{flag_k1 = true;flag_k2 = true;int n;cin >> n;for(int i = 0; i < n; i ++){int a, b, c, d;cin >> a >> b >> c >> d;k1[i] = a;k2[i] = b;num[i][0] = c;num[i][1] = d;}for(int i = 0; i < n; i ++){cnt = 0;node* tree = Init(i);if(cnt == n) //这里没单独寻找根节点，因此采用对每个点假设为根节点，如果建立起来的树节点有n个，则该点就为根节点{midorder(tree); //中序遍历结果，用于后面判断左子树是否满足二叉搜索树的情况Judge(tree); //判断右子树是否为小根堆(k2是否满足条件)break;}}if(!flag_k2) cout << "NO" << endl;else //判断K1，二叉搜索树的中序遍历一定是递减的{for(int i = 0; i < mid.size() - 1; i ++){if(k1[mid[i]] > k1[mid[i + 1]]) //判断中序遍历是不是递减序列，如果不是，则左子树不是二叉搜索树{cout << "NO" << endl;flag_k1 = false;break;}}}if(flag_k1 && flag_k2) cout << "YES" << endl;return 0;
}

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