LeetCode 590: N-ary Tree Postorder Traversal Solution in Python – A Step-by-Step Guide

What Is LeetCode 590: N-ary Tree Postorder Traversal?

In LeetCode 590: N-ary Tree Postorder Traversal, you’re given the root of an N-ary tree, where each node has a value and a list of children (possibly empty), and your task is to return a list of node values in postorder traversal order—visit all children from left to right, then the root. For example, with a tree where the root is 1 with children [5, 6], and 5 has children [3, 2, 4], the result is [3, 2, 4, 5, 6, 1]. This problem builds on LeetCode 145: Binary Tree Postorder Traversal but generalizes to N-ary trees with variable child counts.

Problem Statement

• Input: root (Node)—root of an N-ary tree.
• Output: List[int]—node values in postorder traversal order.
• Rules: Visit all children recursively, then root; each node has a value and list of children.

Constraints

• Number of nodes: 0 to 10⁴
• 0 <= len(node.children) <= 10⁴
• Node values: -10⁴ to 10⁴

Examples

• Input: root = [1,null,3,2,4,null,5,6]
• Output: [5,6,3,2,4,1]
• Tree:
``` 1 /|\ 3 2 4 / \ 5 6 ```
• Postorder: children of 3 [5, 6], 3, then 2, 4, finally 1.
• 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]
• Complex N-ary tree traversal.
• Input: root = []
• Output: []
• Empty tree.

Understanding the Problem: Traversing the N-ary Tree

To solve LeetCode 590: N-ary Tree Postorder Traversal in Python, we need to visit each node in an N-ary tree in postorder—all children left-to-right, then the root—returning a list of values, handling up to 10⁴ nodes efficiently. Unlike binary trees with fixed left and right children, N-ary trees have a dynamic list of children, requiring a flexible traversal approach. A brute-force method manually tracking all nodes could work but is overkill. Instead, a recursive DFS leverages the tree’s structure for O(n) time, while an iterative DFS with a stack offers an alternative. We’ll explore:

• Best Solution (Recursive DFS): O(n) time, O(h) space—fast and elegant (n = nodes, h = height).
• Alternative Solution (Iterative DFS with Stack): O(n) time, O(n) space—thorough and stack-based.

Let’s dive into the recursive solution with a friendly breakdown!

Best Solution: Recursive DFS

Why Recursive DFS Wins

The recursive DFS is the best for LeetCode 590 because it performs the postorder traversal in O(n) time and O(h) space by naturally following the tree’s structure—recursively exploring all children before visiting the root—making it both efficient and intuitive for an N-ary tree’s variable child count. It’s like visiting a family reunion, checking in with each kid’s family first, then greeting the parent—all in a smooth, recursive stroll!

How It Works

Think of this as a tree-visiting guide:

• Step 1: Base Case:
• If root is None, return empty list (no nodes).
• Step 2: Recurse on Children:
• For each child in root.children, recursively traverse and collect results.
• Step 3: Visit Root:
• Append root’s value after all children are processed.
• Step 4: Return Result:
• Combined list of values in postorder.
• Why It Works:
• Recursion handles arbitrary child counts seamlessly.
• Postorder ensures children-first order.

It’s like a postorder family roll call!

Step-by-Step Example

Example: root = [1,null,3,2,4,null,5,6]

• Step 1: Start at root 1, result = [].
• Step 2: Children [3, 2, 4]:
• Node 3:
• Children [5, 6]:
• Node 5: result = [5].
• Node 6: result = [5, 6].
• Visit 3: result = [5, 6, 3].
• Node 2: result = [5, 6, 3, 2].
• Node 4: result = [5, 6, 3, 2, 4].
• Step 3: Visit 1: result = [5, 6, 3, 2, 4, 1].
• Step 4: Return [5, 6, 3, 2, 4, 1].
• Result: [5, 6, 3, 2, 4, 1].

Example: root = [1,null,2]

• Step 1: Start at 1, result = [].
• Step 2: Child [2]:
• Node 2: result = [2].
• Step 3: Visit 1: result = [2, 1].
• Step 4: Return [2, 1].
• Result: [2, 1].

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

Copy codeclass Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children if children is not None else []

class Solution:
    def postorder(self, root: 'Node') -> List[int]:
        # Step 1: Base case - empty tree
        if not root:
            return []

        # Step 2: Initialize result
        result = []

        # Step 3: Recursively traverse children
        for child in root.children:
            result.extend(self.postorder(child))

        # Step 4: Add root value after children
        result.append(root.val)

        # Step 5: Return combined result
        return result

• Lines 2-5: Node class with value and children list.
• Lines 9-11: Base case: return empty list if root is None.
• Line 14: Initialize empty result list.
• Lines 17-18: Recurse on each child, extend result with child’s traversal.
• Line 21: Append root’s value after children.
• Line 24: Return postorder list.
• Time Complexity: O(n)—visit each node once.
• Space Complexity: O(h)—recursion stack (h = tree height).

It’s like a postorder tree maestro!

Alternative Solution: Iterative DFS with Stack

Why an Alternative Approach?

The iterative DFS with stack uses a stack to simulate recursion, running in O(n) time and O(n) space, collecting values in postorder by processing children first, then the root, using a stack to manage traversal order. It’s thorough and avoids recursion stack limits, making it a good alternative for explicit control or large trees.

How It Works

Picture this as a stack-driven explorer:

• Step 1: Initialize stack with root, result list.
• Step 2: While stack not empty:
• Pop node, add value to result (will reverse later).
• Push all children in order (to process left-to-right).
• Step 3: Reverse result to get postorder.
• Step 4: Return result list.

It’s like a stack-guided tree walker!

Step-by-Step Example

Example: root = [1,null,3,2]

• Step 1: Stack = [1], result = [].
• Step 2:
• Pop 1, result = [1], push [3, 2].
• Pop 2, result = [1, 2], no children.
• Pop 3, result = [1, 2, 3], no children.
• Step 3: Reverse result = [3, 2, 1].
• Step 4: Return [3, 2, 1].
• Result: [3, 2, 1].

Code for Iterative Approach

Copy codeclass Solution:
    def postorder(self, root: 'Node') -> List[int]:
        if not root:
            return []

        # Step 1: Initialize stack and result
        stack = [root]
        result = []

        # Step 2: Process stack
        while stack:
            node = stack.pop()
            result.append(node.val)
            # Push children in order (will reverse later)
            for child in node.children:
                stack.append(child)

        # Step 3: Reverse result for postorder
        return result[::-1]

• Lines 4-5: Base case for empty tree.
• Lines 8-9: Start with root in stack, empty result.
• Lines 12-16: Pop node, add value, push children.
• Line 19: Reverse result to achieve postorder.
• Time Complexity: O(n)—each node processed once.
• Space Complexity: O(n)—stack may hold all nodes.

It’s a stack-based postorder pro!

Comparing the Two Solutions

• Recursive DFS (Best):
• Pros: O(n), O(h), simple and elegant.
• Cons: Stack space for deep trees.
• Iterative DFS (Alternative):
• Pros: O(n), O(n), explicit control.
• Cons: More space, requires reversal.

Recursive DFS wins for simplicity!

Additional Examples and Edge Cases

• Empty tree: [].
• Single node: [val].
• Many children: Full traversal.

Recursive DFS handles them all!

Complexity Recap

• Recursive DFS: Time O(n), Space O(h).
• Iterative DFS: Time O(n), Space O(n).

Recursive DFS’s the elegance champ!

Key Takeaways

• Recursive DFS: Tree beauty—learn at Python Basics!
• Iterative DFS: Stack power.
• Trees: Postorder is fun.
• Python: Recurse or stack, your pick!

Final Thoughts: Traverse That Tree!

LeetCode 590: N-ary Tree Postorder Traversal in Python is a delightful tree challenge. Recursive DFS is your fast track, while iterative DFS offers a stack-based twist. Want more? Try LeetCode 145: Binary Tree Postorder Traversal or LeetCode 589: N-ary Tree Preorder Traversal. Ready to explore? Head to Solve LeetCode 590 on LeetCode and list that N-ary tree today!