每日一題 二叉樹節點遍歷的問題

演算法導論p138 10.4-2

題目要求：

給定乙個n節點的二叉樹，寫出乙個o(n)時間的遞迴過程，將概述每個節點的關鍵字輸出。

利用python3來實現。

'''		18

/\12 10
/\ /\
7 4 2 21/5
'''##treenode定義
class
treenode
:def
__init__
(self, x)
: self.val = x
self.left =
none
self.right =
none
def__str__
(self)
:return f'(root: -[left: , right: ])'
##遞迴遍歷
from typing import list
from define.treenode import treenode
class
solution
:# 前序遍歷
defgetallnodes0
(self, root=treenode)
-> list[
int]:if
not self:
return
list()
reslist =
list()
if root:
if root.left:
reslist.extend(solution.getallnodes0(self, root.left)
)if root.right:
reslist.extend(solution.getallnodes0(self, root.right)
)return reslist
# 中序遍歷
defgetallnodes1
(self, root=treenode)
-> list[
int]:if
not self:
return
list()
reslist =
list()
if root:
if root.left:
reslist.extend(solution.getallnodes1(self, root.left)
)if root.right:
reslist.extend(solution.getallnodes1(self, root.right)
)return reslist
# 後序遍歷
defgetallnodes2
(self, root=treenode)
-> list[
int]:if
not self:
return
list()
reslist =
list()
if root:
if root.left:
reslist.extend(solution.getallnodes2(self, root.left)
)if root.right:
reslist.extend(solution.getallnodes2(self, root.right)
)return reslist
if __name__ ==
'__main__'
: root = treenode(18)
root.left = treenode(12)
root.right = treenode(10)
root.left.left = treenode(7)
root.left.right = treenode(4)
root.right.left = treenode(2)
root.right.right = treenode(21)
root.left.right.left = treenode(5)
print
(root)
print
(solution.getallnodes0(self=root, root=root)
)# [18, 12, 7, 4, 5, 10, 2, 21]
print
(solution.getallnodes1(self=root, root=root)
)# [7, 12, 5, 4, 18, 2, 10, 21]
print
(solution.getallnodes2(self=root, root=root)
)# [7, 5, 4, 12, 2, 21, 10, 18]

def

getallnodesbystack
(self, root=treenode)
-> list[
int]:if
not self:
return
stack, reslist =
list()
,[]while stack:
node = stack.pop(
)if node.left:
if node.right:
return reslist

def

getallnodesbyqueue
(self, root=treenode)
-> list[
int]:if
not self:
return
queue, reslist = collections.deque(),
while queue:
node = queue.popleft(
)if node.left:
if node.right:
return reslist

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