求二叉樹中任意兩個結點的距離

實現步驟：

計算跟到第乙個結點的距離；

計算跟到第二個結點的距離；

計算lca；

計算跟到lca結點的距離；

結果為(1) + (2) - 2 * (4)，因為重複計算了兩次的從跟到lca結點的距離；

class

node
(object
):def
__init__
(self
,value=0
):self
.value
=value
self
.left
=self
.right
=none
defget_path_length
(root,n
,path
):if
notroot
:return0if
root
.value==n
:return
path
else
:return
get_path_length
(root
.left,n
,path+1
)orget_path_length
(root
.right,n
,path+1
)def
find_lca
(root,n1
,n2):if
root
isnone
:return
none
ifroot
.value
==n1
orroot
.value
==n2
:return
root
left
=find_lca
(root
.left,n1
,n2)right
=find_lca
(root
.right,n1
,n2)if
left
andright
:return
root
elif
left
:return
left
elif
right
:return
right
else
:return
none
deffind_distance
(root,n1
,n2):x
=get_path_length
(root,n1
,0)y
=get_path_length
(root,n2
,0)lca
=find_lca
(root,n1
,n2).
value
lca_dis
=get_path_length
(root
,lca,0
)return(x
+y)-
2*lca_dis

defget_path_length

(root
,path,k
):# base case handling
ifroot
isnone
:return
false
path.(
root
.value)if
root
.value==k
:return
true
if((
root
.left
!=none
andget_path_length
(root
.left
,path,k
))or
(root
.right
!=none
andget_path_length
(root
.right
,path,k
))):
return
true
# 如果當前結點的值並不是k
path
.pop
()return
false
deffind_distance
(root,n1
,n2):if
root
:# 獲取第乙個結點的路徑（儲存跟結點到i）
path1=
get_path_length
(root
,path1,n1
)# 獲取第二個結點的路徑
path2=
get_path_length
(root
,path2,n2
)# 找到它們的公共祖先i=
0while
i<
len(
path1
)and
i<
len(
path2
):if
path1[i
]!=path2[i
]:breaki=
i+1# 減去重複計算的跟結點到lca部分即為結果
return
(len
(path1)+
len(
path2)-
2*i)
else
:return
0

if__name__

=='__main__'
:root
=node(1
)root
.left
=node(2
)root
.right
=node(3
)root
.left
.left
=node(4
)root
.right
.right
=node(7
)root
.right
.left
=node(6
)root
.left
.right
=node(5
)root
.right
.left
.right
=node(8
)dist
=find_distance
(root,4
,5)print
("distance between node {} & {}: {}"
.format(4
,5,dist
))dist
=find_distance
(root,4
,6)print
("distance between node {} & {}: {}"
.format(4
,6,dist
))dist
=find_distance
(root,3
,4)print
("distance between node {} & {}: {}"
.format(3
,4,dist
))dist
=find_distance
(root,2
,4)print
("distance between node {} & {}: {}"
.format(2
,4,dist
))dist
=find_distance
(root,8
,5)print
("distance between node {} & {}: {}"
.format(8
,5,dist
))

distance between node 4 & 5: 2

distance between node 4 & 6: 4

distance between node 3 & 4: 3

distance between node 2 & 4: 1

distance between node 8 & 5: 5

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