Given an n-ary tree, return the preorder 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: [1,3,5,6,2,4]
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: [1,2,3,6,7,11,14,4,8,12,5,9,13,10]
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 preorder(root):
"""DFS using recursion."""
if not root:
return []
traversal = [root.val] # would come last in postorder
for child in root.children:
traversal.extend(preorder(child))
# !! traversal.extend(preorder(child) for child in root.children) does NOT work
# leads to --> traversal = [w, [x, [y, [z]]]]
return traversal
def preorder(root):
"""alternative version of above solution."""
def helper(node):
if node is None:
return None
nonlocal result
traversal.append(node.val) # would come last in postorder
for child in node.children:
helper(child)
traversal = []
helper(root)
return traversal
Solve it both recursively and iteratively.
def preorder(root):
"""DFS using an iterative approach with a stack."""
if not root:
return []
stack = [root]
traversal = []
while stack:
node = stack.pop()
traversal.append(node.val)
stack += reversed(node.children) # reversed() to ensure left subtree is processed first
# stack.append(reversed(node.children)) doesnt work because reversed() is an iterator
# Instead, we can do:
# stack.extend(reversed(node.children))
# OR
# for child in reversed(node.children):
# stack.append(child)
return traversal