Introduction

In this blog post, we will explore how to solve the LeetCode problem "590" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given the root of 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) 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 number of nodes in the tree is in the range [0, 104]. 0 <= Node.val <= 104 The height of the n-ary tree is less than or equal to 1000. Follow up: Recursive solution is trivial, could you do it iteratively?

Explanation

Here's a solution to the n-ary tree postorder traversal problem, focusing on an efficient iterative approach.

• Iterative Traversal with Stack: The core idea is to use a stack to simulate the recursion. We process each node and its children, ensuring that children are visited before their parent. We employ a stack to keep track of the path from the root to the current node.

• Reverse and Append: Since postorder is left-to-right children then the node, we effectively traverse the tree in a root-right-left order, and then reverse the resulting list to obtain the postorder sequence.

• Handling Null/Separators: In the provided problem's serialization format where children are separated by null, we skip null during traversal to avoid processing it as a node.

• Time & Space Complexity: O(N) time complexity, where N is the number of nodes in the tree. O(H) space complexity, where H is the height of the tree, representing the maximum stack size. In the worst case, H can be equal to N.

Code

    class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children

def postorder(root: 'Node') -> list[int]:
    """
    Performs an iterative postorder traversal of an n-ary tree.

    Args:
        root: The root node of the n-ary tree.

    Returns:
        A list of integers representing the postorder traversal of the tree.
    """
    if not root:
        return []

    stack = [root]
    output = []

    while stack:
        node = stack.pop()
        output.append(node.val)

        if node.children:
            for child in node.children:
                stack.append(child)

    return output[::-1]