使用用户在python中创建n元树

Question

我想用用户提供边缘（格式为uv）和值来创建树。 节点可以具有任意数量的子代。 例如，如果给定的3个节点的值为2 3 4且边的值为1-2和2-3，则树将为2 3 4也不是u <v。并且边是无向的，所以i' ve查找哪一个发生在根附近。

我已经尝试过代码，但是无法制作如下树，但一个节点可以有任意数量的子代

2
 3
  4
   5

以下是代码

class Node : 

# Utility function to create a new tree node 
def __init__(self ,key): 
    self.key = key 
    self.child = []



# Prints the n-ary tree level wise 
def printNodeLevelWise(root): 
if root is None: 
    return

# create a queue and enqueue root to it 
queue = [] 
queue.append(root) 

# Do level order traversal. Two loops are used 
# to make sure that different levels are printed 
# in different lines 
while(len(queue) >0): 

    n = len(queue) 
    while(n > 0) : 

        # Dequeue an item from queue and print it 
        p = queue[0] 
        queue.pop(0) 
        print p.key, 

        # Enqueue all children of the dequeued item 
        for index, value in enumerate(p.child): 
            queue.append(value) 

        n -= 1
    print "" # Seperator between levels 


# test case
t = raw_input()
t=int(t)
while(t > 0):
# number of nodes
n = raw_input()
n=int(n)
# array to keep node value
a = []
nums = raw_input().split()
for i in nums: a.append(int(i))
n = n -1
root = Node(a[0])
i = 1
for j in range(0, n):
    u, v = raw_input().split()
    u=int(u)
    v=int(v)
    if(u == 1):
        root.child.append(Node(a[i]))
    else:
        root.child[u-2].child.append(Node(a[i]))
    i=i+1
t=t-1
printNodeLevelWise(root)

我知道更正应该在root.child[u-2].child.append(Node(a[i]))我希望输出是

2
 3
  4
   5

对于这种情况，但我得到

Traceback (most recent call last):
File "/home/25cd3bbcc1b79793984caf14f50e7550.py", line 52, in <module>
root.child[u-2].child.append(Node(a[i]))
 IndexError: list index out of range

所以我不知道如何纠正它。 请提供正确的代码给我

Answer 1

假设用户给定的边列表为

边= [[3,2]，[3,4]，[4,5]]

根= 2

首先， 我们必须将边列表转换为适当的字典 ，该字典将指示给定边的树结构。 以下是我在StackOverflow上找到的代码。

def createDict(edges, root):
    graph ={}
    for x,y in es:
        graph.setdefault(x,set()).add(y)
        graph.setdefault(y,set()).add(x)
    tree, stack = {}, [s]
    while stack:
        parent = stack.pop()
        children = graph[parent] - tree.keys()
        tree[parent] = list(children)
        stack.extend(children)
    for i in tree.keys():
        if tree[i] == []:    #if the node is leaf node, give it a 0 val
           tree[i].append(0)   
    return tree

输出将是：tree = {2：[3]，3：[4]，4：[5]，5：[0]}

现在，我们将使用Node类将其变为Tree结构 。 这段代码是我写的。

class Node:
     def __init__(self, val):
         self.val = val
         self.children = []  

def createNode(tree, root,b=None, stack=None):
    if stack is None:
        stack = []           #stack to store children values
        root = Node(root)    #root node is created
        b=root               #it is stored in b variable 
    x = root.val             # root.val = 2 for the first time
   if len(tree[x])>0 :       # check if there are children of the node exists or not
      for i in range(len(tree[x])):   #iterate through each child
          y = Node(tree[x][i])        #create Node for every child
          root.children.append(y)     #append the child_node to its parent_node
          stack.append(y)             #store that child_node in stack
          if y.val ==0:     #if the child_node_val = 0 that is the parent = leaf_node 
             stack.pop()    #pop the 0 value from the stack
      if len(stack):        #iterate through each child in stack
          if len(stack)>=2:      #if the stack length >2, pop from bottom-to-top
              p=stack.pop(0)     #store the popped val in p variable
          else:
              p = stack.pop()    #pop the node top_to_bottom
      createNode(tree, p,b,stack)   # pass p to the function as parent_node 
return b                            # return the main root pointer

在这段代码中，b只是一个指向根节点的变量，因此我们可以逐级迭代并打印它。

对于级别顺序打印：

def printLevel(node):
    if node is None:
        return
    queue = []
    queue.append(node)
    while(len(queue)>0):
         n =len(queue)
         while(n>0):
             p = queue[0]
             queue.pop(0)
             if p.val !=0:           #for avoiding the printing of '0'
                print(p.val, end='  ')
                for ind, value in enumerate(p.children):
                      queue.append(value)
             n -= 1
         print(" ")

输出为：

       2
       3
       4
       5  #each level is getting printed

我现在不知道，如何以倾斜的方式打印它）：仍在处理

好吧，我不是Python的专业人士，只是一个初学者。 更正和修改非常受欢迎。