时间:2021-07-01 10:21:17 帮助过:61人阅读
使用类的形式定义二叉树,可读性更好
class BinaryTree: def __init__(self, root): self.key = root self.left_child = None self.right_child = None def insert_left(self, new_node): if self.left_child == None: self.left_child = BinaryTree(new_node) else: t = BinaryTree(new_node) t.left_child = self.left_child self.left_child = t def insert_right(self, new_node): if self.right_child == None: self.right_child = BinaryTree(new_node) else: t = BinaryTree(new_node) t.right_child = self.right_child self.right_child = t def get_right_child(self): return self.right_child def get_left_child(self): return self.left_child def set_root_val(self, obj): self.key = obj def get_root_val(self): return self.key r = BinaryTree('a') print(r.get_root_val()) print(r.get_left_child()) r.insert_left('b') print(r.get_left_child()) print(r.get_left_child().get_root_val()) r.insert_right('c') print(r.get_right_child()) print(r.get_right_child().get_root_val()) r.get_right_child().set_root_val('hello') print(r.get_right_child().get_root_val())
Python进行二叉树遍历
需求:
python代码实现二叉树的:
1. 前序遍历,打印出遍历结果
2. 中序遍历,打印出遍历结果
3. 后序遍历,打印出遍历结果
4. 按树的level遍历,打印出遍历结果
5. 结点的下一层如果没有子节点,以‘N'代替
方法:
使用defaultdict或者namedtuple表示二叉树
使用StringIO方法,遍历时写入结果,最后打印出结果
打印结点值时,如果为空,StringIO()写入‘N '
采用递归访问子节点
代码
#!/usr/bin/env python3 # -*- coding: utf-8 -*- # test tree as below: ''' 1 / \ / \ / \ / \ 2 3 / \ / \ / \ / \ 4 5 6 N / \ / \ / \ 7 N N N 8 9 / \ / \ / \ N N N N N N ''' from collections import namedtuple from io import StringIO #define the node structure Node = namedtuple('Node', ['data', 'left', 'right']) #initialize the tree tree = Node(1, Node(2, Node(4, Node(7, None, None), None), Node(5, None, None)), Node(3, Node(6, Node(8, None, None), Node(9, None, None)), None)) #read and write str in memory output = StringIO() #read the node and write the node's value #if node is None, substitute with 'N ' def visitor(node): if node is not None: output.write('%i ' % node.data) else: output.write('N ') #traversal the tree with different order def traversal(node, order): if node is None: visitor(node) else: op = { 'N': lambda: visitor(node), 'L': lambda: traversal(node.left, order), 'R': lambda: traversal(node.right, order), } for x in order: op[x]() #traversal the tree level by level def traversal_level_by_level(node): if node is not None: current_level = [node] while current_level: next_level = list() for n in current_level: if type(n) is str: output.write('N ') else: output.write('%i ' % n.data) if n.left is not None: next_level.append(n.left) else: next_level.append('N') if n.right is not None: next_level.append(n.right) else: next_level.append('N ') output.write('\n') current_level = next_level if __name__ == '__main__': for order in ['NLR', 'LNR', 'LRN']: if order == 'NLR': output.write('this is preorder traversal:') traversal(tree, order) output.write('\n') elif order == 'LNR': output.write('this is inorder traversal:') traversal(tree, order) output.write('\n') else: output.write('this is postorder traversal:') traversal(tree, order) output.write('\n') output.write('traversal level by level as below:'+'\n') traversal_level_by_level(tree) print(output.getvalue())
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