PAT A1155 Heap Paths (30point(s))

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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))

One thing for sure is that all the keys along any path from the root to a leaf in a max/min heap must be in non-increasing/non-decreasing order.

Your job is to check every path in a given complete binary tree, in order to tell if it is a heap or not.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (1<N≤1,000), the number of keys in the tree. Then the next line 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, first print all the paths from the root to the leaves. Each path occupies a line, with all the numbers separated by a space, and no extra space at the beginning or the end of the line. The paths must be printed in the following order: for each node in the tree, all the paths in its right subtree must be printed before those in its left subtree.

Finally 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.

分析

给出结点个数和一颗完全二叉树的层次遍历序列,从右往左,输出所有从根节点到叶子结点的路径,然后判断该二叉树是大顶堆还是小顶堆,如果都不是,就输出Not Heap。

解答

由题意可知,遍历顺序为右->中->左,可以参考中序遍历,编写相关代码。当访问到叶子结点时,说明已经生成一条从根节点到叶子结点的路径,因为不知道该序列是递增的,还是递减的,可以计算每两个相邻键的差值,如果差值恒为正或负则必定为递增或递减,如果既有正又有负,说明该二叉树一定不是大顶堆或小顶堆。

#include<iostream>
#include<vector>
#include<set>

#define NOTHEAP 0
#define MAXHEAP 1
#define MINHEAP 2

using namespace std;

int keysNum;
vector<int> path, nodes;
set<int> flags;
void preOrderReverse(int id)
{
	path.push_back(nodes[id]);
	if (2 * id > keysNum) { // 叶子结点
		bool sub[2] = { false, false };
		printf("%d", path[0]);
		for (int i = 1; i < path.size(); i++) {
			printf(" %d", path[i]);
			if (path[i] < path[i - 1]) sub[0] = true;
			else if(path[i] > path[i - 1]) sub[1] = true;
		}
		if (sub[0] && sub[1]) flags.insert(NOTHEAP);
		else if (sub[0]) flags.insert(MAXHEAP);
		else if (sub[1]) flags.insert(MINHEAP);
		printf("\n");
		path.pop_back();
		return;
	}
        // 有右孩子才访问左子树,否则会多输出一条路径
	if (id * 2 + 1 <= keysNum) preOrderReverse(id * 2 + 1);
	preOrderReverse(id * 2);
	path.pop_back();
}
int main()
{
	cin >> keysNum;
	nodes.resize(keysNum + 1);
	for (int i = 0; i < keysNum; i++) cin >> nodes[i + 1];
	preOrderReverse(1);
	if (flags.size() != 1 || *flags.begin() == NOTHEAP) printf("Not Heap");
	else if (*flags.begin() == MINHEAP) printf("Min Heap");
	else if (*flags.begin() == MAXHEAP) printf("Max Heap");
	return 0;
}