Python实现二叉树的遍历

1 class Node:  
2     def __init__(self,value=None,left=None,right=None):  
3         self.value=value  
4         self.left=left    #左子树
5         self.right=right  #右子树

 1 def preTraverse(root):  
 2     ‘‘‘
 3     前序遍历
 4     ‘‘‘
 5     if root==None:  
 6         return  
 7     print(root.value)  
 8     preTraverse(root.left)  
 9     preTraverse(root.right)  
10 
11 def midTraverse(root): 
12     ‘‘‘
13     中序遍历
14     ‘‘‘
15     if root==None:  
16         return  
17     midTraverse(root.left)  
18     print(root.value)  
19     midTraverse(root.right)  
20   
21 def afterTraverse(root):  
22     ‘‘‘
23     后序遍历
24     ‘‘‘
25     if root==None:  
26         return  
27     afterTraverse(root.left)  
28     afterTraverse(root.right)  
29     print(root.value)  

 1 if __name__==‘__main__‘:
 2     root=Node(‘D‘,Node(‘B‘,Node(‘A‘),Node(‘C‘)),Node(‘E‘,right=Node(‘G‘,Node(‘F‘))))
 3     print(‘前序遍历：‘)
 4     preTraverse(root)
 5     print(‘\n‘)
 6     print(‘中序遍历：‘)
 7     midTraverse(root)
 8     print(‘\n‘)
 9     print(‘后序遍历：‘)
10     afterTraverse(root)
11     print(‘\n‘)

前序遍历：
D
B
A
C
E
G
F


中序遍历：
A
B
C
D
E
F
G


后序遍历：
A
C
B
F
G
E
D

 1 preList = list(‘12473568‘)
 2 midList = list(‘47215386‘)
 3 afterList = []
 4 
 5 def findTree(preList, midList, afterList):
 6     if len(preList) == 0:
 7         return
 8     if len(preList) == 1:
 9         afterList.append(preList[0])
10         return
11     root = preList[0]
12     n = midList.index(root)
13     findTree(preList[1:n + 1], midList[:n], afterList)
14     findTree(preList[n + 1:], midList[n + 1:], afterList)
15     afterList.append(root)

[‘7‘, ‘4‘, ‘2‘, ‘5‘, ‘8‘, ‘6‘, ‘3‘, ‘1‘]

1 如果以上面的前序：DBACEGF和中序：ABCDEFG，得到的结果为：
2 [‘A‘, ‘C‘, ‘B‘, ‘F‘, ‘G‘, ‘E‘, ‘D‘]