python数据结构与算法——二叉树结构与遍历方法

1 # 二叉树结点类型
2 class BTNode:
3     def __init__(self,value,lft=None,rgt=None):
4         self.value = value
5         self.lft = lft          # 结点左分支 BTNode
6         self.rgt = rgt          # 结点右分支 BTNode

 1 class BinTree():
 2     def __init__(self):
 3         self.root = None    # 创建一个空的二叉树
 4         
 5     def isEmpty(self):      # 判断二叉树是否为空
 6         if self.root is None: return True
 7         else: return False
 8     
 9     def makeBT(self,bt,L=None,R=None):       # 从当前结点创建二叉树
10         bt.lft = L
11         bt.rgt = R
12     
13     def returnBTdict(self):              # 返回二叉树的字典模式
14         if self.isEmpty(): 
15             return None
16         def rec(bt=None,R=True):
17             if R==True:
18                 bt = self.root
19                 return {‘root‘:{‘value‘:bt.value,"L":rec(bt.lft,False),
20                                                 "R":rec(bt.rgt,False)} }
21             else:
22                 if bt==None:
23                     return None
24                 else:
25                     return {"value":bt.value,
26                             "L":rec(bt.lft,False) if bt.lft != None else None,
27                             "R":rec(bt.rgt,False) if bt.rgt != None else None}
28             return None
29         return rec()
30             
31     def __repr__(self):             # 将二叉树结构打印为字典结构
32         return str(self.returnBTdict())

 1     def printT_VLR(self,bt=None,rec_count = 0):     # 输出二叉树结构（先序遍历）
 2         # rec_count 用于计算递归深度 以便输出最后的换行符
 3         """
 4         # 先序遍历
 5         # 访问结点，遍历左子树，如果左子树为空，则遍历右子树，
 6         # 如果右子树为空，则向上走到一个可以向右走的结点，继续该过程
 7         preorder(t):
 8             if t:
 9                 print t.value
10                 preorder t.L
11                 preorder t.R
12         """
13         if bt==None: 
14             bt = self.root
15             print bt.value,
16         btL, btR = bt.lft, bt.rgt
17         if btL != None:
18             print btL.value,;   rec_count += 1;     self.printT_VLR(btL,rec_count);     rec_count -= 1
19         if btR != None:
20             print btR.value,;   rec_count += 1;     self.printT_VLR(btR,rec_count);     rec_count -= 1
21         if rec_count == 0:
22             print "\n"
23     

 1     def printT_LVR(self,bt=None):
 2         """
 3         # 中序遍历
 4         # 从根开始，一直走向左下方，直到无结点可以走则停下，访问该节点
 5         # 然后走向右下方到结点，继续走向左下方：如果结点无右孩子，则向上走回父亲结点
 6         inorder(t):
 7             inorder(t.L)
 8             print t.value
 9             inorder(t.R)
10         """
11         if bt==None:
12             bt = self.root
13         btL, btR = bt.lft, bt.rgt
14         if btL != None:
15             self.printT_LVR(btL)
16         
17         print bt.value,
18 
19         if btR != None:
20             self.printT_LVR(btR)

 1     def printT_LRV(self,bt=None):
 2         """
 3         # 后序遍历
 4         inorder(t):
 5             inorder(t.L)
 6             inorder(t.R)
 7             print t.value
 8         """
 9         if bt==None:
10             bt = self.root
11         btL, btR = bt.lft, bt.rgt
12         if btL != None:
13             self.printT_LRV(btL)
14         if btR != None:
15             self.printT_LRV(btR)
16         print bt.value,

 1     def printT_levelorder(self):
 2         """
 3         层序遍历 采用队列的遍历操作
 4         第一次访问根，在访问根的左孩子，接着访问根的有孩子，然后下一层
 5         自左向右一一访问同层的结点
 6         """
 7         btdict = self.returnBTdict()
 8         q = []
 9         q.append(btdict[‘root‘])
10         while q:
11             tn = q.pop(0)   # 从队列中弹出一个结点(也是一个字典)
12             print tn["value"],
13             if tn["L"]!=None:
14                 q.append(tn["L"])
15             if tn["R"]!=None:
16                 q.append(tn["R"])
17     

 1 def test():
 2     bt = BinTree()
 3 
 4 #     btns = [BTNode(v) for v in "+*E*D/CAB"]     # 层序输入
 5 #     bt.root = btns[0]
 6 #     bt.makeBT(btns[0], L=btns[1], R=btns[2])
 7 #     bt.makeBT(btns[1], L=btns[3], R=btns[4])
 8 #     bt.makeBT(btns[3], L=btns[5], R=btns[6])
 9 #     bt.makeBT(btns[5], L=btns[7], R=btns[8])
10     
11     btns = [BTNode(v) for v in [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]]
12     bt.root = btns[0]
13     bt.makeBT(btns[0], L=btns[1], R=btns[2])
14     bt.makeBT(btns[1], L=btns[3], R=btns[4])
15     bt.makeBT(btns[2], L=btns[5], R=btns[6])
16     bt.makeBT(btns[3], L=btns[7], R=btns[8])
17     bt.makeBT(btns[4], L=btns[9], R=btns[10])
18     bt.makeBT(btns[5], L=btns[11], R=btns[12])
19     bt.makeBT(btns[6], L=btns[13], R=btns[14])

{‘root‘: {‘R‘: {‘R‘: {‘R‘: {‘R‘: None, ‘L‘: None, ‘value‘: 15}, ‘L‘: {‘R‘: None, ‘L‘: None, ‘value‘: 14}, ‘value‘: 7}, ‘L‘: {‘R‘: {‘R‘: None, ‘L‘: None, ‘value‘: 13}, ‘L‘: {‘R‘: None, ‘L‘: None, ‘value‘: 12}, ‘value‘: 6}, ‘value‘: 3}, ‘L‘: {‘R‘: {‘R‘: {‘R‘: None, ‘L‘: None, ‘value‘: 11}, ‘L‘: {‘R‘: None, ‘L‘: None, ‘value‘: 10}, ‘value‘: 5}, ‘L‘: {‘R‘: {‘R‘: None, ‘L‘: None, ‘value‘: 9}, ‘L‘: {‘R‘: None, ‘L‘: None, ‘value‘: 8}, ‘value‘: 4}, ‘value‘: 2}, ‘value‘: 1}}