N-ary Tree Postorder Traversal

LeetCode 776 | Difficulty: Easy

Easy

Problem Description

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, 10^4]`.

- `0 <= Node.val <= 10^4`

- 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?

Topics: Stack, Tree, Depth-First Search

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Approach

Tree DFS

Traverse the tree recursively (or with a stack). At each node, decide: what information do I need from the left/right subtrees? Process: go left → go right → combine results. Consider preorder, inorder, or postorder traversal based on when you need to process the node.

When to use

Path problems, subtree properties, tree structure manipulation.

Stack

Use a stack (LIFO) to track elements that need future processing. Process elements when a "trigger" condition is met (e.g., finding a smaller/larger element). Monotonic stack maintains elements in sorted order for next greater/smaller element problems.

When to use

Matching brackets, next greater element, evaluating expressions, backtracking history.

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Solutions

Solution 1: C# (Best: 380 ms)

Metric	Value
Runtime	380 ms
Memory	34.7 MB
Date	2020-01-25

Solution

/*
// Definition for a Node.
public class Node {
    public int val;
    public IList<Node> children;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, IList<Node> _children) {
        val = _val;
        children = _children;
    }
}
*/
public class Solution {
    public IList<int> Postorder(Node root) {
        Stack<Node> s1 = new Stack<Node>();
        Stack<Node> s2 = new Stack<Node>();
        List<int> result = new List<int>();
        if(root==null) return result;
        s1.Push(root);
        while(s1.Count != 0)
        {
            var popped = s1.Pop();
            s2.Push(popped);
            foreach (var child in popped.children)
            {
                s1.Push(child);
            }
        }
        while(s2.Count!=0)
        {
            result.Add(s2.Pop().val);
        }
        return result;
    }
}

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Complexity Analysis

Approach	Time	Space
Tree Traversal	$O(n)$	$O(h)$
Stack	$O(n)$	$O(n)$

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Interview Tips

Key Points

• Start by clarifying edge cases: empty input, single element, all duplicates.
• Consider: "What information do I need from each subtree?" — this defines your recursive return value.
• Think about what triggers a pop: is it finding a match, or finding a smaller/larger element?