N-ary Tree Postorder Traversal - Practice Coding Problems

N-ary Tree Postorder Traversal - Problem

Welcome to the fascinating world of N-ary trees! Unlike binary trees that have at most 2 children per node, N-ary trees can have any number of children - making them perfect for representing hierarchical structures like file systems, organizational charts, or family trees.

Your mission is to perform a postorder traversal of an N-ary tree and return the values in the correct order. In postorder traversal, we visit all children nodes first, then the parent node - think of it as exploring all rooms in a house before examining the main entrance.

What you're given:

The root of an N-ary tree
Each node can have 0 to N children

What you need to return:

A list containing all node values in postorder traversal sequence

Note: The tree input uses level-order serialization where each group of children is separated by null values.

Input & Output

example_1.py — Basic N-ary Tree

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

› Output: [5,6,3,2,4,1]

💡 Note: The tree has root 1 with children [3,2,4]. Node 3 has children [5,6]. In postorder, we visit 5,6 first, then 3, then 2, then 4, and finally root 1.

example_2.py — Complex Tree

$ 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]

💡 Note: A more complex tree structure where each subtree is processed in postorder before moving to the parent node.

example_3.py — Single Node

$ Input: root = [1]

› Output: [1]

💡 Note: Edge case with just one node - the postorder traversal simply returns that single node's value.

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?

Visualization

Tap to expand

Understanding the Visualization

Start the Journey

Begin at the family patriarch (root node)

Explore Descendants

Visit all children and their descendants first

Meet Ancestors

Only after meeting all descendants, meet the current ancestor

Complete Lineage

Continue until the entire family tree is explored

Key Takeaway

🎯 Key Insight: Postorder traversal ensures we process all descendants before their ancestors - perfect for operations requiring bottom-up processing like calculating directory sizes or expression evaluation.

Asked in

G Google 45 a Amazon 38 ⊞ Microsoft 32 f Meta 28 Apple 22

The optimal solution uses an iterative stack approach to avoid recursion overhead. By performing a modified preorder traversal (visiting children right-to-left) and then reversing the result, we achieve the correct postorder sequence in O(n) time with minimal space usage. This approach is stack-overflow safe and preferred in interviews.

Common Approaches

Approach	Time	Space	Notes
✓ Naive Recursive Approach	O(n)	O(h)	Simple recursive solution without optimizations
Iterative Stack Approach (Optimal)	O(n)	O(h)	Use an explicit stack to simulate recursion and achieve postorder traversal

Naive Recursive Approach — Algorithm Steps

Start from the root node
For each node, recursively traverse all its children
After visiting all children, add the current node's value to result
Return the complete result list

Visualization

Tap to expand

Step-by-Step Walkthrough

Start at Root

Begin traversal at the root node

Recurse on Children

Recursively visit all children before processing current node

Process Node

Add current node value to result after all children are processed

Return Result

Complete traversal returns postorder sequence

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

// Definition for a Node.
struct Node {
    int val;
    int numChildren;
    struct Node** children;
};

int* result;
int resultSize = 0;

void dfs(struct Node* node) {
    if (!node) return;
    
    // First, visit all children
    for (int i = 0; i < node->numChildren; i++) {
        dfs(node->children[i]);
    }
    
    // Then, add current node's value
    result[resultSize++] = node->val;
}

int* postorder(struct Node* root, int* returnSize) {
    *returnSize = 0;
    if (!root) return NULL;
    
    result = (int*)malloc(10000 * sizeof(int));
    resultSize = 0;
    
    dfs(root);
    *returnSize = resultSize;
    return result;
}

int main() {
    // Parse input and call solution
    printf("[]\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

We visit each node exactly once, where n is the total number of nodes

✓ Linear Growth

Space Complexity

O(h)

Recursion depth equals tree height h, plus O(n) for result storage

✓ Linear Space

67.2K Views

Medium Frequency

~12 min Avg. Time

1.8K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Naive Recursive Approach — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler