[英]Preorder Binary Tree traversal Recursive method
I was trying to understand implementation of Binary Tree traversal (PreOrder).我试图了解二叉树遍历(PreOrder)的实现。 The non recursive approach was fine, but I am totally lost while trying to understand the recursive approach.
非递归方法很好,但是在尝试理解递归方法时我完全迷失了。
Code:代码:
def preorder_print(self, start, traversal): """Root->Left->Right"""
if start:
traversal += (str(start.value) + "-")
traversal = self.preorder_print(start.left, traversal)
traversal = self.preorder_print(start.right, traversal)
return traversal
Binary Tree二叉树
8
/ \
4 5
/ \ \
2 1 6
My understanding is while reaching Node 2(8-4-2), left of that node 2 is None.我的理解是在到达节点 2(8-4-2)时,节点 2 的左侧是无。 So
if start:
condition would fail.所以
if start:
条件会失败。
Below are my questions.以下是我的问题。
My understanding on recursion is poor, kindly help!我对递归的理解很差,请帮助!
Watch this, see if this helps:看这个,看看这是否有帮助:
class Node(object):
def __init__(self, value):
self.value = value
self.left = None
self.right = None
def addleft(self,value):
self.left = Node(value)
def addright(self,value):
self.right = Node(value)
def preorder_print(self, start, traversal='', depth=0):
print( " "*depth, start.value if start else "None" )
if start:
traversal += (str(start.value) + "-")
print(' '*depth, "check left")
traversal = self.preorder_print(start.left, traversal, depth+1)
print(' '*depth, "check right")
traversal = self.preorder_print(start.right, traversal, depth+1)
return traversal
base = Node(8)
base.addleft(4)
base.left.addleft(2)
base.left.addright(1)
base.addright(5)
base.right.addright(6)
print( base.preorder_print( base ) )
Output: Output:
8
check left
4
check left
2
check left
None
check right
None
check right
1
check left
None
check right
None
check right
5
check left
None
check right
6
check left
None
check right
None
8-4-2-1-5-6-
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