python - 递归后序遍历到 Python 中的列表？

Question

大家好——我是一个编程新手，这里有以下非常简单的代码：

def postorder(T):
    if T != None:
        postorder(T.left)
        postorder(T.right)
        print T.data,

我想要的不是打印遍历，而是让函数将该信息存储在数组或类似的东西中，这样我就可以将该信息用于其他事情

Answer 1

你可以做：

def postorder(tree):
    data = []

    def recurse(node)
        if not node:
            return
        recurse(node.left)
        recurse(node.right)
        data.append(node.data)

    recurse(tree)
    return data

内部函数recurse负责遍历树并将数据自动添加到data.

Answer 2

你可以传入一个可调用对象，或者你可以编写一个生成器：

def postorder(T):
    if T != None:
        postorder(T.left)
        postorder(T.right)
        print T.data,

def postorder_func(T, f):
    if T != None:
        postorder_func(T.left, f)
        postorder_func(T.right, f)
        f(T.data)

def postorder_gen(T):
    if T != None:
        for i in postorder_gen(T.left):
            yield i
        for i in postorder_gen(T.right):
            yield i
        yield T.data


class T:
    def __init__(self, left = None, data = None, right = None):
        self.left = left
        self.data = data
        self.right = right

one_half = T(data=.5)
three_halves = T(data=1.5)
one = T(one_half, 1, three_halves)
three = T(data=3)
root = T(one, 2, three)

postorder(root)
print ""

a = []
postorder_func(root, a.append)
print a

a = list(postorder_gen(root))
print a

---

或者，单线解决方案：

def postorder_one(T):
    return [] if T is None else postorder_one(T.left)+[T.data]+postorder_one(T.right)