Given an n-ary tree, return the postorder traversal of its nodes' values.
Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).
Follow up:
Recursive solution is trivial, could you do it iteratively?
Example 1:
Input: root = [1,null,3,2,4,null,5,6] Output: [5,6,3,2,4,1]
Example 2:
Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14] Output: [2,6,14,11,7,3,12,8,4,13,9,10,5,1]
Constraints:
1000
[0, 10^4]
Source
# Definition for a Node.
class Node:
def __init__(self, val=None, children=None):
self.val = val
self.children = children # children = list
def postorder(root):
"""DFS using recursion."""
if not root:
return []
traversal = []
for child in root.children:
traversal.extend(postorder(child))
traversal.append(root.val) # would come first in preorder
return traversal
def postorder(root):
"""alternative version of above solution."""
def helper(node):
if node is None:
return None
nonlocal traversal
for child in node.children:
helper(child)
traversal.append(node.val) # would come first in preorder
traversal = []
helper(root)
return traversal
Solve it both recursively and iteratively.
def postorder(root):
"""DFS using an iterative approach with a stack."""
if root is None:
return []
stack = [root]
traversal = []
while stack:
node = stack.pop()
traversal.append(node.val)
stack += node.children # Option 1
# stack.extend(node.children) # Option 2
# for child in node.children: # Option 3
# stack.append(child)
return reversed(traversal)