堆与堆排序的问题

[lihujun101]

def _siftdown(self, ndx):
        left = 2 * ndx + 1
        right = 2 * ndx + 2
        # determine which node contains the larger value
        largest = ndx
        if (left < self._count and     # 有左孩子
                self._elements[left] >= self._elements[largest] and
                self._elements[left] >= self._elements[right]):  # 原书这个地方没写实际上找的未必是largest
            largest = left
        elif right < self._count and self._elements[right] >= self._elements[largest]:
            largest = right
        if largest != ndx:
            self._elements[ndx], self._elements[largest] = self._elements[largest], self._elements[ndx]
            self._siftdown(largest)

若self.elements = [4,3,2,1,0]这种结构的话，这种会报错，是否应该先判断右孩子是否存在，然后再判断左孩子？

  def _siftdown(self, ndx):
        left = 2 * ndx + 1
        right = 2 * ndx + 2
        # determine which node contains the larger value
        largest = ndx
        if (right < self._count and  # 有右孩子
                self._elements[right] >= self._elements[largest] and
                self._elements[left] <= self._elements[right]):  # 原书这个地方没写实际上找的未必是largest
            largest = right
        elif left < self._count and self._elements[left] >= self._elements[largest]:
            largest = left
        if largest != ndx:
            self._elements[ndx], self._elements[largest] = self._elements[largest], self._elements[ndx]
            self._siftdown(largest)

[PegasusWang]

有 test case 么？ self.extract() 调用 _siftdown 之前 self._count 执行了减去1的操作， 这里 right = 2*ndx * 2 . 似乎不会越界。