N-ary Tree Postorder Traversal

Problem Statement

In this problem, we are given the root of an n-ary tree and need to conduct a postorder traversal on it, returning the values of its nodes in the sequence they are visited. An n-ary tree, as opposed to a binary tree, is a tree in which each node can have more than two children. The tree's input is given in a serialized form representing its level order traversal; each set of children is delineated by a null value indicating the end of children for the current node. Our task is to understand and manipulate this structure to execute a postorder traversal, which involves visiting the node's children first before the node itself.

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, 104].
• 0 <= Node.val <= 104
• The height of the n-ary tree is less than or equal to 1000.

Approach and Intuition

The challenge requires a postorder traversal of an n-ary tree, framed in terms of serialized level order data. Let's identify the steps for a postorder traversal:

1. Traverse all the children nodes first.
2. Visit the root node last.

Given the provided serialization format, here's how you might conceptualize underlying operations:

• For the first example:

  • Create the n-ary tree from serialized data [1,null,3,2,4,null,5,6].
  • Here, 1 is the root.
  • 1 has three children [3, 2, 4], and 3 has two children [5, 6].
  • Applying postorder traversal: Traverse children of 3 -> 3 -> children of 2 -> 2 -> children of 4 -> 4 -> 1. Resulting in [5, 6, 3, 2, 4, 1].

• For the second example:

  • The tree is larger with deeper levels, but the approach remains the same.
  • Begin from the deepest and left-most child, traverse up and ensure each parent is visited after its children.

Postorder traversal ensures that we process from the bottom-up in a tree, making it particularly useful in scenarios where operations on children must precede operations on parents, such as in certain tree dynamic programming problems or when freeing memory of a tree starting from the leaves.

Solutions

class Solution {
public:
    vector<int> traversePostorder(Node* root) {
        vector<int> output;
        if (!root) return output;
    
        stack<NodeVisitPair> traversalStack;
        traversalStack.push(NodeVisitPair(root, false));
    
        while (!traversalStack.empty()) {
            NodeVisitPair currentNode = traversalStack.top();
            traversalStack.pop();
    
            if (currentNode.visited) {
                output.push_back(currentNode.node->val);
            } else {
                currentNode.visited = true;
                traversalStack.push(currentNode);
    
                vector<Node*>& childNodes = currentNode.node->children;
                for (int i = childNodes.size() - 1; i >= 0; --i) {
                    traversalStack.push(NodeVisitPair(childNodes[i], false));
                }
            }
        }
    
        return output;
    }
    
private:
    struct NodeVisitPair {
        Node* node;
        bool visited;
    
        NodeVisitPair(Node* nodePtr, bool isVisited) : node(nodePtr), visited(isVisited) {}
    };
};