无法使用面向对象的prog生成fibonacci树。在python中。

本文介绍了无法使用面向对象的prog生成fibonacci树。在python中。的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 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() 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. 这篇关于无法使用面向对象的prog生成fibonacci树。在python中。的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！