[英]Depth-first search in Python never returns True
When searching for a '3' in a tree that clearly contains a 3, the code will go into the correct "if statement", but does not return True.在明确包含 3 的树中搜索 '3' 时,代码将进入正确的“if 语句”,但不返回 True。 Why is that?
这是为什么?
class Node:
def __init__(self):
self.value = None
self.leftChild = None
self.rightChild = None
def dfs(root, sought):
print "root", root.value
print "sought", sought.value
if root.value == sought.value:
print "inside root = sought"
return True
elif root.leftChild is not None and root.rightChild is not None:
dfs(root.leftChild, sought)
dfs(root.rightChild, sought)
return False
The tree that I created with the sought value s is as follows:我用寻求的值 s 创建的树如下:
n1 = Node()
n2 = Node()
n3 = Node()
n4 = Node()
n5 = Node()
n1.leftChild = n2
n1.rightChild = n5
n2.leftChild = n3
n2.rightChild = n4
n1.value=1
n2.value=2
n3.value=3
n4.value=4
n5.value=5
s = Node()
s.value=3
But the output is befuddling.但输出令人眼花缭乱。 I have a hunch the lack of
return True
has something to do with Python's recursive nature?我有一种预感,缺少
return True
与 Python 的递归性质有关吗?
> dfs(n1,s)
> root 1
> sought 3
> root 2
> sought 3
> root 3
> sought 3
> inside root = sought
> root 4
> sought 3
> root 5
> sought 3
> False
You need to return the results from your recursive calls:您需要返回递归调用的结果:
res_a = dfs(root.leftChild, sought)
res_b = dfs(root.rightChild, sought)
return res_a or res_b
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