Question

我有一个二叉树，每个节点存储一段数据（self.pred）。我想编写一个函数，对于树中的每个叶子，返回访问的节点中的数据列表（这是一个lambda函数）以到达该叶子。

例如，我想要在这棵树上调用函数：

    A 
   / \
 B    C
/ \  / \
1  2 3  4

返回：

returned_list = [
[A, B, 1]
[A, B, 2]
[A, C, 3]
[A, C, 4]
]

为了使思考更复杂，按照右分支到达一条数据必须返回存储为该节点数据的lambda函数的逆。

例如，A'为not(A)，必须返回以下列表：

returned_list = [
[A, B, 1]
[A, B', 2]
[A', C, 3]
[A', C', 4]
]

这是我到目前为止所做的：

class Node:
    def __init__(self, pred):
        self.parent = None
        self.left = None
        self.right = None
        self.pred = pred

def build_tree(pred_choices, k):
    if k == 0:
        return Node(get_pred())
    else:
        node = Node(get_pred())
        node.left = build_tree(pred_choices, k-1)
        node.right = build_tree(pred_choices, k-1)
        return node

def get_paths(root, cur_path, all_paths):
    if (root.left == None and root.right == None): # If leaf, append current path
        all_paths.append(cur_path)
    else:
        get_paths(root.left, cur_path.append(root.pred), all_paths)
        get_paths(root.right, cur_path.append(lambda x: not(root.pred(x)), all_paths)
    return all_paths

def auto_tree_build(pred_choices, k): 
    root = build_tree(pred_choices, k)
    all_paths = get_paths(root, [], [])
    return all_paths

但上面的代码不起作用，我不明白它的输出。有人可以帮助我让上面的代码执行所描述的行为吗？

Answer 1

我会使用递归。

我稍微更改了Node课程，但只要每个from copy import copy from collections import namedtuple # NodePoint, to distinguish between A and A' NP = namedtuple('NP', ['node', 'right']) class Node(object): def __init__(self, name, left=None, right=None): self.name = name self.left = left self.right = right d = Node('1') e = Node('2') f = Node('3') g = Node('4') b = Node('B', d, e) c = Node('C', f, g) a = Node('A', b, c) def get_routes(node, route=None): route = route or [] # If this node has children, clone the route, so we can process # both the right and left child separately. if node.right: right_route = copy(route) # Process the main (left) route. Pass the route on to the left # child, which may return multiple routes. route.append(NP(node, False)) routes = get_routes(node.left, route) if node.left else [route] # If there is a right child, process that as well. Add the route # results. Note that NP.right is set to True, to indicate a right path. if node.right: right_route.append(NP(node, True)) right_routes = get_routes(node.right, right_route) routes.extend(right_routes) # Pass the results back return routes routes = get_routes(a) # print out the results. for route in routes: names = [] for np in route: name = '{0}{1}'.format(np.node.name, "'" if np.right else '') names.append(name) print names # ['A', 'B', '1'] # ['A', "B'", '2'] # ["A'", 'C', '3'] # ["A'", "C'", '4']存储左右儿童，您就可以设计它。

//$dbname= 'your connection to database';
$result = mysqli_query($dbname, 'SELECT txtpath,txtname FROM txtdocs WHERE subject="'.$ref.'"');
$count = mysqli_num_rows($result);
//iterate over all the rows
while($row = mysqli_fetch_assoc($result)){
    echo $row['txtpath'] . ': ' .$row['txtname']. '<br/>';
}

获取导致Python中的叶子的所有路径的列表

1 个答案: