问题
An array of integers of size n can be converted into a heap by adjusting the heaps rooted at each internal node of the complete binary tree starting at the node ⌊(n−1)/2⌋ and doing this adjustment up to the root node (root node is at index 0) in the order ⌊(n−1)/2, ⌊(n−3)/2⌋, ....., 0.
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I know, it's a Build Heap procedure and takes O(n) time, but can someone please make me visualise by taking a array with small value of n to show how things are working?
回答1:
Let's do an example with the array [7, 3, 9, 1, 2, 4, 8, 5, 6, 0]. The tree structure is:
7
3 9
1 2 4 8
5 6 0
There are 10 items in the heap. So, starting at index (n/2)-1 (the first non-leaf node), we see that the value 2 is greater than its one child. We swap, giving:
7
3 9
1 0 4 8
5 6 2
Next, 1 is smaller than its children, so we leave it alone.
9 is larger than both of its children. The rule is that you swap it with its smallest child, giving:
7
3 4
1 0 9 8
5 6 2
3 is larger than both of its children, so swap it with 0. It's also larger than 2, so we do another swap. The result is:
7
0 4
1 2 9 8
5 6 3
Finally, 7 is larger than 0, so we swap it, placing 0 at the root. 7 is also larger than its two children, so we swap it with 1. And 7 is larger than 5 and 6, so we swap it to the leaf level. The result is:
0
1 4
5 2 9 8
7 6 3
Which is a valid min-heap.
来源：https://stackoverflow.com/questions/50947087/build-heap-procedure