589. N-ary Tree Preorder Traversal
https://leetcode.com/problems/n-ary-tree-preorder-traversal/description/
这道题与二叉树的遍历很类似，同样可以使用递归和迭代两种方法来解。递归非常好实现，迭代需要注意使用栈结构和children入栈的顺序。
递归代码如下：
"""
# Definition for a Node.
class Node(object):
def __init__(self, val, children):
self.val = val
self.children = children
"""
class Solution(object):
def preorder(self, root):
"""
:type root: Node
:rtype: List[int]
"""
result = []
self.helper(root, result)
return result
def helper(self, root, result):
if not root:
return
result.append(root.val)
for child in root.children:
self.helper(child, result)
迭代代码如下：
"""
# Definition for a Node.
class Node(object):
def __init__(self, val, children):
self.val = val
self.children = children
"""
class Solution(object):
def preorder(self, root):
"""
:type root: Node
:rtype: List[int]
"""
result = []
if not root:
return result
stack = [root]
while stack:
node = stack.pop()
result.append(node.val)
stack += [child for child in node.children[::-1] if child]
return result
590. N-ary Tree Postorder Traversal
https://leetcode.com/problems/n-ary-tree-postorder-traversal/description/
解法与上一题基本相同，不再赘述。
递归代码如下：
"""
# Definition for a Node.
class Node(object):
def __init__(self, val, children):
self.val = val
self.children = children
"""
class Solution(object):
def postorder(self, root):
"""
:type root: Node
:rtype: List[int]
"""
result = []
self.helper(root, result)
return result
def helper(self, root, result):
if not root:
return
for child in root.children:
self.helper(child, result)
result.append(root.val)
迭代代码如下：
"""
# Definition for a Node.
class Node(object):
def __init__(self, val, children):
self.val = val
self.children = children
"""
class Solution(object):
def postorder(self, root):
"""
:type root: Node
:rtype: List[int]
"""
result = []
if not root:
return result
stack = [root]
while stack:
node = stack.pop()
result.append(node.val)
stack += node.children
return result[::-1]