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algorithms-python/algorithms/tree/path_sum.py at master · Mu-L/algorithms-python · GitHub
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algorithms-python
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algorithms
/
tree
/
path_sum.py
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algorithms-python
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algorithms
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tree
/
path_sum.py
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"""
Given a binary tree and a sum, determine if the tree has a root-to-leaf
path such that adding up all the values along the path equals the given sum.
For example:
Given the below binary tree and sum = 22,
5
/
\
4 8
/ /
\
11 13 4
/ \
\
7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
"""
def
has_path_sum
(
root
,
sum
):
"""
:type root: TreeNode
:type sum: int
:rtype: bool
"""
if
root
is
None
:
return
False
if
root
.
left
is
None
and
root
.
right
is
None
and
root
.
val
==
sum
:
return
True
sum
-=
root
.
val
return
has_path_sum
(
root
.
left
,
sum
)
or
has_path_sum
(
root
.
right
,
sum
)
# DFS with stack
def
has_path_sum2
(
root
,
sum
):
if
root
is
None
:
return
False
stack
=
[(
root
,
root
.
val
)]
while
stack
:
node
,
val
=
stack
.
pop
()
if
node
.
left
is
None
and
node
.
right
is
None
:
if
val
==
sum
:
return
True
if
node
.
left
is
not
None
:
stack
.
append
((
node
.
left
,
val
+
node
.
left
.
val
))
if
node
.
right
is
not
None
:
stack
.
append
((
node
.
right
,
val
+
node
.
right
.
val
))
return
False
# BFS with queue
def
has_path_sum3
(
root
,
sum
):
if
root
is
None
:
return
False
queue
=
[(
root
,
sum
-
root
.
val
)]
while
queue
:
node
,
val
=
queue
.
pop
(
0
)
# popleft
if
node
.
left
is
None
and
node
.
right
is
None
:
if
val
==
0
:
return
True
if
node
.
left
is
not
None
:
queue
.
append
((
node
.
left
,
val
-
node
.
left
.
val
))
if
node
.
right
is
not
None
:
queue
.
append
((
node
.
right
,
val
-
node
.
right
.
val
))
return
False
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