1147. Heaps (30) 堆

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本文介绍了一种算法，用于判断给定的完全二叉树是否满足最大堆或最小堆的性质，并输出其后序遍历序列。通过递归方式检查每个节点与其子节点的关系来确定整棵树是否符合堆的定义。

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1147. Heaps (30)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

In computer science, a heap is a specialized tree-based data structure that satisfies the heap property: if P is a parent node of C, then the key (the value) of P is either greater than or equal to (in a max heap) or less than or equal to (in a min heap) the key of C. A common implementation of a heap is the binary heap, in which the tree is a complete binary tree. (Quoted from Wikipedia at https://en.wikipedia.org/wiki/Heap_(data_structure))

Your job is to tell if a given complete binary tree is a heap.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers: M (<= 100), the number of trees to be tested; and N (1 < N <= 1000), the number of keys in each tree, respectively. Then M lines follow, each contains N distinct integer keys (all in the range of int), which gives the level order traversal sequence of a complete binary tree.

Output Specification:

For each given tree, print in a line "Max Heap" if it is a max heap, or "Min Heap" for a min heap, or "Not Heap" if it is not a heap at all. Then in the next line print the tree's postorder traversal sequence. All the numbers are separated by a space, and there must no extra space at the beginning or the end of the line.

Sample Input:

3 8
98 72 86 60 65 12 23 50
8 38 25 58 52 82 70 60
10 28 15 12 34 9 8 56

Sample Output:

Max Heap
50 60 65 72 12 23 86 98
Min Heap
60 58 52 38 82 70 25 8
Not Heap
56 12 34 28 9 8 15 10

给出一个层序遍历的树，判断是不是最大堆或者最小堆，最后按后序遍历输出。

输入数据的时候建立二叉树。给出的是层序遍历，那么最后的树除了右下肯定是充满了的，用一个list存储没有右子树的节点地址，每次把新节点接在list最前面的节点上，如果最前面的节点左右子树都有了就pop，即可建立二叉树。

之后思路很清楚的，递归判断，通过返回值判断子树类型，再结合自身节点情况继续向上返回即可，判断过程中可以以后序顺序收集数据，最后输出即可。（后久没做二叉树居然忘记了后序是怎么遍历的了。。。

#include <cstdio>
#include <cstdlib>
#include <list>
#include <vector>

using namespace std;

typedef struct Node *TREE;
typedef struct Node {
	int data;
	TREE l;
	TREE r;
}Node;

int M, N;
int sz[100010];
Node *T;
vector<int>v1;

int isHeap(TREE t) {
	if ((!t->l) && (!t->r)) {
		v1.push_back(t->data);
		return 3;
	}
	if (!t->r) {
		int f = isHeap(t->l);
		v1.push_back(t->data);
		return t->data > t->l->data ? 1 : 2;
	}
	if (t->data > t->l->data&&t->data > t->r->data) {
		int ll = isHeap(t->l);
		int rr = isHeap(t->r);
		v1.push_back(t->data);
		if (ll == 0 || rr == 0 || ll == 2 || rr == 2)return 0;
		return 1;
	}
	else if (t->data < t->l->data&&t->data < t->r->data) {
		int ll = isHeap(t->l);
		int rr = isHeap(t->r);
		v1.push_back(t->data);
		if (ll == 0 || rr == 0 || ll == 1 || rr == 1)return 0;
		return 2;
	}
	else {
		int ll = isHeap(t->l);
		int rr = isHeap(t->r);
		v1.push_back(t->data);
		return 0;
	}
}

void pf() {
	int n = v1.size();
	for (int i = 0; i < n; i++) {
		printf("%d%c", v1[i], i == n - 1 ? '\n' : ' ');
	}
}

void checkk() {
	v1.clear();
	int f = T->data > T->l->data ? 1 : 2;//0:Not Heap  1:Max Heap  2:Min Heap  3:All
	f = isHeap(T);
	if (f == 1) {
		printf("Max Heap\n");
	}
	else if (f == 2) {
		printf("Min Heap\n");
	}
	else {
		printf("Not Heap\n");
	}
	pf();
}

int main() {
	scanf("%d %d\n", &M, &N);
	for (int i = 0; i < M; i++) {
		int t;
		list<TREE>ls;
		TREE rot = NULL;
		for (int j = 0; j < N; j++) {
			scanf("%d", &t);
			TREE tt = (TREE)malloc(sizeof(Node));
			tt->data = t;
			tt->l = NULL;
			tt->r = NULL;
			ls.push_back(tt);
			if (!rot) {
				rot = tt;
			}
			else {
				if ((ls.front())->l) {
					(ls.front())->r = tt;
					ls.pop_front();
				}
				else {
					(ls.front())->l = tt;
				}
			}
		}
		T = rot;
		checkk();
	}
	return 0;
}