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灰色功能接受物体作为参数吗?
原标题:Python function accepting an object as a parameter?

因此,我有:

class Tree(object):
    def__init__(self):
        self.right = None
        self.left = None
        self.val = None

I populate the tree with stuff in my code.

我愿写一下表格的功能。

def mult(newVal, node)
    newVal = diff(node.left, newVal, var)
    newVal.append( * )
    newval.append(next(node.right))
    newVal.append( + )
    newVal = diff(node.left, newVal, var)
    newVal.append( * )
    newVal.append(next(node.left))

下面只是一种功能,即渗透到树中的下一个 no子上,扩散是一种可再生功能,其形式如下:

def diff(node, newVal, var):
...
...
    elif(node.val ==  * ):
        newVal = diff(node.left, newVal, var)
        newVal.append( * )
        newval.append(next(node.right))
        newVal.append( + )
        newVal = diff(node.left, newVal, var)
        newVal.append( * )
        newVal.append(next(node.left))
...
...

我的“主要”

    node = Tree()
    newEquation = []
    pos = 0
    pos, newTree = buildTree(node, equation, pos)
    newEquation = diff(newTree, newEquation, variable)
    newEquation =   .join(newEquation)
    print newEquation

我将如何写上<代码>mult()功能,接受树冠作为参数,然后把<代码>diff(再次运作?

最佳回答

甲型六氯环己烷没有打字,因此,如果你说:

mult(newEquation, node)

这应当只是罚款。

问题回答

暂无回答




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