问题
用递归方式遍历二叉树
思路说明
遍历二叉树的方法有广度优先和深度优先两类,下面阐述的是深度优先。
以下图的二叉树为例:
先定义三个符号标记:
- 访问结点本身(N)
- 遍历该结点的左子树(L)
- 遍历该结点的右子树(R)
有四种方式:
- 前序遍历(PreorderTraversal,NLR):先访问根结点,然后遍历其左右子树
- 中序遍历(InorderTraversal,LNR):先访问左子树,然后访问根节点,再访问右子树
- 后序遍历(PostorderTraversal,LRN):先访问左右子树,再访问根结点
- 层序遍历(levelorderTraversal):按照从上到下的层顺序访问
上面的数,按照以上四种方式遍历,得到的结果依次是:
- preorder: 1 2 4 7 5 3 6 8 9
- inorder: 7 4 2 5 1 8 6 9 3
- postorder: 7 4 5 2 8 9 6 3 1
- level-order: 1 2 3 4 5 6 7 8 9
下面用递归的方式,解决此题。
解决(Python)
1. #! /usr/bin/env python2. #coding:utf-83. 4. from collections import namedtuple5. from sys import stdout6. 7. Node = namedtuple('Node', 'data, left, right')8. tree = Node(1,9. Node(2,10. Node(4,11. Node(7, None, None),12. None),13. Node(5, None, None)),14. Node(3,15. Node(6,16. Node(8, None, None),17. Node(9, None, None)),18. None))19. 20. 21. #前序(pre-order,NLR)22. 23. def preorder(node):24. if node is not None:25. print node.data,26. preorder(node.left)27. preorder(node.right)28. 29. 30. #中序(in-order,LNR)31. 32. def inorder(node):33. if node is not None:34. inorder(node.left)35. print node.data,36. inorder(node.right)37. 38. 39. #后序(post-order,LRN)40. 41. def postorder(node):42. if node is not None:43. postorder(node.left)44. postorder(node.right)45. print node.data,46. 47. 48. #层序(level-order)49. 50. def levelorder(node, more=None):51. if node is not None:52. if more is None:53. more = []54. more += [node.left, node.right]55. print node.data,56. if more:57. levelorder(more[0], more[1:])58. 59. if __name__=="__main__"60. print ' preorder: ',61. preorder(tree)62. print '\t\n inorder: ',63. inorder(tree)64. print '\t\n postorder: ',65. postorder(tree)66. print '\t\nlevelorder: ',67. levelorder(tree)68. print '\n'
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