LeetCode 589: N-ary Tree Preorder Traversal Solution in Python – A Step-by-Step Guide

What Is LeetCode 589: N-ary Tree Preorder Traversal?

In LeetCode 589: N-ary Tree Preorder 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 preorder traversal order—visit the root first, then recursively traverse all children from left to right. For example, with a tree where the root is 1 with children [5, 6], and 5 has children [3, 2, 4], the result is [1, 5, 3, 2, 4, 6]. This problem builds on LeetCode 144: Binary Tree Preorder 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 preorder traversal order.
• Rules: Visit root, then all children recursively; 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: [1,3,5,6,2,4]
• Tree:
``` 1 /|\ 3 2 4 / \ 5 6 ```
• Preorder: root (1), then children [3, 2, 4], with 3’s children [5, 6].
• 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: [1,2,3,6,7,11,14,4,8,12,5,9,10,13]
• Complex N-ary tree traversal.
• Input: root = []
• Output: []
• Empty tree.

Understanding the Problem: Traversing the N-ary Tree

To solve LeetCode 589: N-ary Tree Preorder Traversal in Python, we need to visit each node in an N-ary tree in preorder—root first, then all children left-to-right—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 unnecessary. Instead, a recursive DFS leverages the tree’s natural 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 589 because it performs the preorder traversal in O(n) time and O(h) space by naturally following the tree’s structure—visiting the root, then recursively exploring each child—making it both efficient and intuitive for an N-ary tree’s variable child count. It’s like visiting a family reunion, starting with the head, then greeting each kid’s family in turn—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: Visit Root:
• Add root’s value to result list.
• Step 3: Recurse on Children:
• For each child in root.children, recursively traverse and append results.
• Step 4: Return Result:
• Combined list of values in preorder.
• Why It Works:
• Recursion handles arbitrary child counts seamlessly.
• Preorder ensures root-first order.

It’s like a preorder 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: Visit 1, result = [1].
• Step 3: Children [3, 2, 4]:
• Node 3: result = [1, 3].
• Children [5, 6]:
• Node 5: result = [1, 3, 5].
• Node 6: result = [1, 3, 5, 6].
• Node 2: result = [1, 3, 5, 6, 2].
• Node 4: result = [1, 3, 5, 6, 2, 4].
• Step 4: Return [1, 3, 5, 6, 2, 4].
• Result: [1, 3, 5, 6, 2, 4].

Example: root = [1,null,2]

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

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

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

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

        # Step 2: Initialize result with root value
        result = [root.val]

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

        # Step 4: Return combined result
        return result

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

It’s like a tree-traversal 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 preorder by processing nodes and their children in reverse 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.
• Push children in reverse order (to process left-to-right).
• Step 3: 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 [2, 3].
• Pop 3, result = [1, 3], no children.
• Pop 2, result = [1, 3, 2], no children.
• Step 3: Return [1, 3, 2].
• Result: [1, 3, 2].

Code for Iterative Approach

class Solution:
    def preorder(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 reverse to process left-to-right
            for child in reversed(node.children):
                stack.append(child)

        # Step 3: Return result
        return result

• 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 in reverse.
• Line 19: Return preorder list.
• Time Complexity: O(n)—each node processed once.
• Space Complexity: O(n)—stack may hold all nodes.

It’s a stack-based traverser!

Comparing the Two Solutions

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

Recursive DFS wins for elegance!

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 simplicity champ!

Key Takeaways

• Recursive DFS: Tree elegance—learn at Python Basics!
• Iterative DFS: Stack control.
• Trees: Traversals are fun.
• Python: Recurse or stack, your pick!

Final Thoughts: Traverse That Tree!

LeetCode 589: N-ary Tree Preorder 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 144: Binary Tree Preorder Traversal or LeetCode 590: N-ary Tree Postorder Traversal. Ready to explore? Head to Solve LeetCode 589 on LeetCode and traverse that N-ary tree today!