Notebook

590. N-ary Tree Postorder Traversal

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:

The height of the n-ary tree is less than or equal to 1000
The total number of nodes is between [0, 10^4]

Source

In [4]:

# Definition for a Node.
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children  # children = list

Code

In [5]:

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

In [6]:

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

Follow up:

Solve it both recursively and iteratively.

Code

In [7]:

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)