对于任一结点。能够按某种次序运行三个操作:
用来表示顺序,即,前序NLR/中序LNR/后序LRN.
- 訪问结点本身(N)
- 遍历该结点的左子树(L)
- 遍历该结点的右子树(R)
用来表示顺序,即,前序NLR/中序LNR/后序LRN.
以下我们用namedtuple来表达树,而通杀的遍历函数带一个order參数,仅仅要我们把指定顺序传进去就可以实现相应的遍历.
#coding=utf-8 ''' 1 / / / 2 3 / / 4 5 6 / / 7 8 9 ''' from collections import namedtuple from sys import stdout Node = namedtuple('Node', ['data','left','right']) 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)) def visitor(i): stdout.write("%i "%i) def traversal(node, order):#这个是主角,通杀函数 if not node:return op = { 'N':lambda:visitor(node.data), 'L':lambda:traversal(node.left, order), 'R':lambda:traversal(node.right, order), } for x in order: op[x]() for order in ['NLR', 'LNR', 'LRN']: traversal(tree, order) print