Question

我正在尝试在python中创建一个树，我希望以后能够进行广泛的首次遍历。但是，当我尝试遍历节点的子节点（即根节点）时，它似乎找不到任何子节点。但是，如果我尝试单独访问子进程（root.children [1]），它是有效的。我很确定我做的事情很傻，但我不确定在哪里。这是我的代码：

class Component:
    def __init__(self, compName, originName, connected = False): #removed to decrease code size
    def __str__(self):
        return "\n\tComponent Name: {0}\n\tOrigin Name: {1}\n\tConnected: {2}\n".format(self.name, self.origin, str(self.connected));

class Node:
    def __init__(self, component, parent):
        self.children = [];    
    def add_node(self, node):       
        self.children.append(node);
        #...left out some broken code that is commented out...#
    def __str__(self):
        return "{0}".format(str(self.component));

class Tree:
    def __init__(self):
    def add_component(self, component, parent = None):
        if self.root is None:
            self.root = Node(component, parent);
        else:
            self.root.add_node(Node(component, parent));

def breadth_first(node):
    result = [];
    queue = [];
    queue.append(node);

    while queue:
        node = queue.pop(0);
        result.append(node);
        plog("Adding to result\n");
        plog(str(node));

        if node in node.children:
            for child in node.children:
                if child not in result and child not in queue:
                    queue.append(child);
        else:
            plog("NO CHILDREN\n");  // this is being displayed
    return result;

def main(ScriptArgument, oDesktop):
    component = Component("Laminate", "B7_TX");
    tree = Tree();
    tree.add_component(component);

    component = Component("B7", "B7_TX");
    tree.add_component(component, "Laminate");

    result = breadth_first(tree.root);
    for x in result:
        plog(str(x) + "\n");

这是我得到的输出：

Adding to result           # output from breadth-first

Component Name: Laminate   #output from breadth-first
Origin Name: B7_TX
Connected: False
NO CHILDREN                # this is indicating that the child was not added properly, I believe

Component Name: Laminate   # output from the loop in main
Origin Name: B7_TX
Connected: False

Answer 1

为什么要检查父项是否在具有以下行的子项列表中？

if node in node.children:

我会这样做：

[...]
while queue:
    node = queue.pop(0);
    result.append(node);
    plog("Adding to result\n");
    plog(str(node));

    for child in node.children:
        if child not in result and child not in queue:
            queue.append(child);
return result;

Answer 2

我的＆＃34;如果＆＃34;检查错了。我正在检查根（或当前）节点是否在其自己的子列表中，这是错误的。我只是改变它来检查是否有任何子节点（如果node.children :)并且它有效。

使用Python将子节点添加到树中

2 个答案: