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.
Sample Input 1:
8
98 72 86 60 65 12 23 50
Sample Output 1:
98 86 23
98 86 12
98 72 65
98 72 60 50
Max Heap
Sample Input 2:
8
8 38 25 58 52 82 70 60
Sample Output 2:
8 25 70
8 25 82
8 38 52
8 38 58 60
Min Heap
Sample Input 3:
8
10 28 15 12 34 9 8 56
Sample Output 3:
10 15 8
10 15 9
10 28 34
10 28 12 56
Not Heap
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注意:
完全二叉树的定义:若设二叉树的深度为h,除第h层外,其他各层(1~h-1)的节点数都达到最大个数,第h层所有节点都连续集中在最左边(就是说可能会有只有左子树没有右子树的情况,按照宽度优先遍历的时候搜索到空节点时树一定已经遍历完)。完全二叉树是由满二叉树而引出来的。对于深度为k的,由n个节点的二叉树,当且仅当其每一个节点的深度都与深度为k的二叉树中编号从1到n的节点一一对应称之为完全二叉树。
题目大意:
给出一颗完全二叉树,打印从根节点开始到所有叶节点的路径,打印的顺序先右后坐,及先序遍历的镜像。然后判断是否是堆。
分析:
1.深度打印所有的路径(从右向左,及先序的镜像),vector保存一条路径上的所有节点,通过push和pop实现回溯,维护路径,index<=n是对只有左节点没有右节点的情况的特判。
2.判断是否为堆:从第二个节点开始遍历,如果比父节点小,就不是小顶堆,如果比父节点大,就不是大顶堆。
#include<stdio.h> #include<vector> #pragma warning(disable:4996) #define maxn 1004 using namespace std; int n; int he[maxn]; vector<int>road; void dfs(int index) { if (index * 2 > n) { if (index <= n) { for (int i = 0; i < road.size(); i++) { printf("%d%s", he[road[i]], i == road.size()-1 ? "\n" : " "); } }//对只有左子树没有右子树的特判 } else { road.push_back(index*2+1); dfs(index * 2 + 1); road.pop_back(); road.push_back(index * 2); dfs(index * 2); road.pop_back(); } } int main() { scanf("%d",&n); for (int i = 1; i <= n; i++) { scanf("%d", &he[i]); } road.push_back(1); dfs(1); int ismax = 1, ismin = 1; for (int i = 2; i <= n; i++) { if (he[i / 2] > he[i])ismin = 0; if (he[i / 2] < he[i])ismax = 0; } printf("%s",ismin ==1?"Min Heap":(ismax == 1?"Max Heap":"Not Heap")); getchar(); getchar(); return 0; }
原文地址:https://www.cnblogs.com/zxzmnh/p/11629929.html