如何在Python中返回随机二叉树的所有可能路径

我有以下形式的随机二叉树

12

13、14

29、26、89

每个节点都有两个子节点,即(12-(13,14),13-(29,26),14-(26,89)).在这里,我需要以[[12,13,29],[12,13,26],[12,14,26],[12,14,89]]的形式返回所有可能的路径.我尝试了以下代码.我有更新列表的问题.提前致谢.

class Tree:

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

    def __str_(self):
        return '%s' % self.data

def makeList(tree, path =[]):
    if(tree != None):
        path.append(tree.data)
        if tree.left:
            path.append(makeList(tree.left, path))
        if tree.right:
            path.append(makeList(tree.left, path))

    return path

根=树(12)

root.left =树(13)

root.right =树(14)

root.right.left =树(26)

root.left.right =树(26)

root.left.left =树(29)

root.right.right =树(86)

x = makeList(root)

解决方法:

我不知道如何使用memoized recursion解决它.但是我仍然发布我的答案,因为它可能部分解决了您的问题.

def makeList(tree):
    paths = []
    if not (tree.left or tree.right):
        return [[tree.data]]
    if tree.left:
        paths.extend([[tree.data] + child for child in makeList(tree.left)])
    if tree.right:
        paths.extend([[tree.data] + child for child in makeList(tree.right)])
    return paths

标签：random,binary-tree,tree,python
来源： https://codeday.me/bug/20191121/2050460.html