python中有堆吗

"ki<=k2i,ki<=k2i+1;或ki>=k2i,ki>=k2i+1.（i=1,2,…,[n/2])"

import heapq
1.heapq.heappush(heap,item) #heap为定义堆，item 增加的元素;
eg.
 heap=[]
 heapq.heappush(heap, 2)
2.heapq.heapify(list) #将列表转换为堆
eg.
 list=[5,8,0,3,6,7,9,1,4,2]
 heapq.heapify(list) 
3.heapq.heappop(heap) #删除最小的值
eg.
 heap=[2, 4, 3, 5, 7, 8, 9, 6]
 heapq.heappop(heap) ---->heap=[3, 4, 5, 7, 9, 6, 8]
4.heapq.heapreplace(heap, item) #删除最小元素值，添加新的元素值
eg.
 heap=[2, 4, 3, 5, 7, 8, 9, 6]
 heapq.heapreplace(heap, 11) ------>heap=[2, 3, 4, 6, 8, 5, 7, 9, 11]
5.heapq.heappushpop(heap, item) #首判断添加元素值与堆的第一个元素值对比，如果大于则删除最小元素，然后添加新的元素值，否则不更改堆
eg.
 条件：item >heap[0]
 heap=[2, 4, 3, 5, 7, 8, 9, 6]
 heapq.heappushpop(heap, 9)---->heap=[3, 4, 5, 6, 8, 9, 9, 7]
 条件：item
 heap=[2, 4, 3, 5, 7, 8, 9, 6]
 heapq.heappushpop(heap, 9)---->heap=[2, 4, 3, 5, 7, 8, 9, 6]
6.heapq.merge(...) #将多个堆合并
7.heapq.nlargest (n, heap) #查询堆中的最大元素，n表示查询元素个数
eg.
 heap=[2, 3, 5, 6, 4, 8, 7, 9]
 heapq.nlargest (1, heap)--->[9]
8.heapq.nsmallest(n, heap) #查询堆中的最小元素，n表示查询元素个数
eg.
 heap=[2, 3, 5, 6, 4, 8, 7, 9]
 heapq.nlargest (1, heap)--->[2]