How do you solve N-ary Tree Postorder Traversal on LeetCode?

N-ary Tree Postorder Traversal (LeetCode #590) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Postorder traversal visits children before their parent nodes.

What is the time complexity of N-ary Tree Postorder Traversal?

The optimal solution for N-ary Tree Postorder Traversal runs in O(n) time and uses O(n) space. The time complexity is linear because we visit each node exactly once. The space complexity is linear due to the stack storing nodes.

What is the brute force solution for N-ary Tree Postorder Traversal?

The brute force approach for N-ary Tree Postorder Traversal is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves recursively visiting each node and collecting their values in postorder. This means we first visit all children of a node before visiting the node itself. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve N-ary Tree Postorder Traversal?

N-ary Tree Postorder Traversal uses the following concepts: Stack, Tree, Depth-First Search. The recommended patterns to study are: Depth-First Search, Stack.

What are the interview tips for N-ary Tree Postorder Traversal?

Explain your thought process clearly while coding. Consider edge cases, like an empty tree. Practice both recursive and iterative solutions.

What are common mistakes when solving N-ary Tree Postorder Traversal?

Forgetting to reverse the result list in the iterative approach. Not handling null nodes properly.

#590

N-ary Tree Postorder Traversal

Easy

StackTreeDepth-First SearchDepth-First SearchStack

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses an iterative approach with a stack to simulate the postorder traversal without recursion. This is efficient and avoids the overhead of recursive calls.

⚙️

Algorithm

6 steps

1Step 1: Initialize an empty stack and push the root node onto it.
2Step 2: Create an empty list to store the result.
3Step 3: While the stack is not empty, pop a node from the stack.
4Step 4: Add the node's value to the result list.
5Step 5: Push all the children of the popped node onto the stack.
6Step 6: Reverse the result list before returning it.

solution.py17 lines

1# Full working Python code
2class Node:
3    def __init__(self, val=None, children=None):
4        self.val = val
5        self.children = children if children is not None else []
6
7class Solution:
8    def postorder(self, root):
9        if not root:
10            return []
11        stack = [root]
12        result = []
13        while stack:
14            node = stack.pop()
15            result.append(node.val)
16            stack.extend(node.children)
17        return result[::-1]

ℹ

Complexity note: The time complexity is linear because we visit each node exactly once. The space complexity is linear due to the stack storing nodes.

1Postorder traversal visits children before their parent nodes.
2Using a stack allows us to handle the traversal iteratively.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.