建议写成递归形式，我用Python似的代码演示一下：

def get_current(current): if not hasleaf(current):return current pos = get_min(current) # 返回 0：current, -1: left, 1: right swap(current, pos) return get_current(current)

get_min的逻辑也不是很复杂。

建议阅读一下PriorityQueue的源码 , 内部有一个siftUp()和siftDown()两个函数, 一个是将元素浮上来, 一个是将元素沉下去, 如果要删除任意节点, 那么也就是把末尾的节点补到删除元素的位置, 然后沉下去, 再浮上来就可以了.

这个是我前几天复习数据结构随手写的, 没有经过测试, 不过主体的逻辑还算正确

public class Heap> { private T[] heap ; private int size ; @SuppressWarnings("unchecked") public Heap(int N) { heap = (T[])new Object[N] ; size = 0 ; } public boolean isEmpty() { return size != 0 ; } //你要实现的那个函数 public void delete(int pos) { if(pos == size-1) { heap[pos] = null ; return ; } heap[pos] = heap[size-1] ; size-- ; sink(pos , heap[pos]); swim(pos , heap[pos]); } public void insert(T t) { swim(size , t) ; size++ ; } private void swim(int index , T t) { while (index > 0) { int parent = (index-1)>>>1 ; T e = heap[parent] ; if(t.compareTo(e) >= 0) break ; heap[parent] = t ; index = parent ; } heap[index] = t ; } public T deleteMin() { T t = heap[0] ; T e = heap[--size] ; heap[size+1] = null ; if(size != 0) sink(0 , e) ; return t ; } private void sink(int index , T t) { while (index<<1+1 < size) { int min = index<<1+1 ; if(min+1 < size && heap[min].compareTo(heap[min+1]) > 0) min++ ; if(heap[min].compareTo(t) > 0) break ; heap[index] = heap[min] ; index = min ; } heap[index] = t ; } }

public void delete(int pos) { Heap[pos] = Heap[size]; size--; int current = pos; while (hasLeaf(current)) { if (hasDoubleLeaf(current) && Heap[current] > Heap[swapNode = getMinChild(current)]) { swap(swapNode, current); current = swapNode; } else if (Heap[current] > Heap[leftChild(current)]) { swap(current, leftChild(current)); current = leftChild(current); } else{ break; } } } } private int getMinChild(int current){ return Heap[leftChild(current)] < Heap[rightChild(current)]? leftChild(current):rightChild(current); }

因为本身的业务逻辑就在那里,所以想从减少if分支的话其实很难做到,而且在各个分支里面的逻辑也不是很复杂,也没有必须要到抽象成接口的程度.个人观点,欢迎交流