无法使用面向对象的prog生成fibonacci树。在python中。 [英] Unable to generate fibonacci tree with object oriented prog. In python.

问题描述

I am a starter & want to integrate dfs code with Fibonacci series generating code. The Fibonacci code too runs as dfs, with calls made from left to right.
The integration is incomplete still. 


I have two issues :
(i) Unable to update 'path' correctly in fib(), as the output is not correctly depicting that.

(ii) Stated in fib() function below, as comment.

-----

Have one more issue that is concerned with program's working:

(iii) On modifying line #16 to: stack = root = stack[1:]; get the same output as before.



 
              
    import sys
    count = 0
    root_counter = 0
    path=1
    inf = -1
    node_counter = 0
    root =0

    def get_depth_first_nodes(root):
        nodes = []
        stack = [root]
        while stack:
            cur_node = stack[0]
            stack = stack[1:]
            nodes.append(cur_node)        
            for child in cur_node.get_rev_children():
                stack.insert(0, child)
        return nodes

    def node_counter_inc():
         global node_counter
         node_counter = node_counter + 1

    class Node(object):
        def __init__(self, id_,path):
            self.id = node_counter_inc() 
            self.children = []
            self.val = inf #On instantiation, val = -1, filled bottom up; 
                           #except for leaf nodes
            self.path = path                
            
        def add_child(self, node):
            self.children.append(node)     
        def get_children(self):
            return self.children        
        def get_rev_children(self):
            children = self.children[:]
            children.reverse()
            return children        
    
    def fib(n, level, val, path):
        global count, root_counter, root
        print('count :', count, 'n:', n, 'dfs-path:', path)
        count += 1
        if n == 0 or n == 1:
            path = path+1
            return n
        if root_counter == 0:
            root = Node(n, path)
            root_counter = 1
        else:
            #cur_node.add_child(Node(n, path)) -- discarded for next(new) line
            root.add_child(Node(n, path))   
        tmp = fib(n-1, level + 1,inf, path) + fib(n-2, level + 1,inf,path+1)
        #Issue 2: Need update node's val field with tmp.  
        #So, need suitable functions in Node() class for that.
        print('tmp:', tmp, 'level', level)
        return tmp

    def test_depth_first_nodes():
         fib(n,0,-1,1)  
         node_list = get_depth_first_nodes(root)
         for node in node_list:
             print(str(node))

    if __name__ == "__main__":
         n = int(input("Enter value of 'n': ")) 
         test_depth_first_nodes() 

< b>我尝试了什么：



除了上述尝试之外，还浪费地尝试在fib（）中使用cur_node（而不是root）。但这并没有成功，因为它被视为整数。我对上述程序进行了以下更改：1。将其初始化为'0'作为g.v.; 2.在两个函数中将其声明为g.v.：fib（）和get_depth_first_nodes（）; 3.注释第53行而不是第52行。

What I have tried:

Apart from the above attempt, also wastefully tried to use cur_node (instead of root) in fib(). But that did not work out as it was taken as integer. I made following changes to the above program : 1. initializing it to '0' as a g.v.; 2. declaring it as g.v.in both functions: fib() and get_depth_first_nodes(); 3. commenting line #53 instead of line #52.

推荐答案

转到 Python教程 - Python 3.7.2文档 [ ^ ]，其中包含有关如何创建斐波纳契程序的出色解释。

Go to The Python Tutorial — Python 3.7.2 documentation[^], which contains excellent explanations of how to create a Fibonacci program.

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