N-ary Tree Postorder Traversal

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) {}
    };
};

This solution implements a postorder traversal for an N-ary tree using C++. The tree structure allows each node to have any number of child nodes, rather than being limited to binary trees.

Here's how the solution works:

• Begin with the creation of a vector<int> to store the traversal results. If the root is null, simply return the empty vector, indicating there's nothing to traverse.
• Use a stack of NodeVisitPair, a custom struct that holds a node and a boolean to keep track of whether a node has been visited. This approach helps manage backtracking in traversal without using recursion.
• Iterate using a while loop as long as the stack is not empty. In each iteration, check the top node of the stack:
  • If the node has not been marked as visited, mark it as visited and push it back onto the stack. Then, push all of its children onto the stack in reverse order. This reverse order ensures that children are considered in the correct sequence later because stacks follow a Last In First Out (LIFO) approach.
  • If the node has already been visited, it means that all children of that node have been processed. Add this node's value to the output vector.
• Once the stack is empty, return the output vector containing the node values in postorder.

This non-recursive method efficiently handles trees with deeply nested children or with a very high number of nodes, potentially reducing the overhead of system stack usage as seen in recursive approaches. This method is particularly useful in environments with limited stack size.

class Solution {

    public List<Integer> postorderTraversal(Node rootNode) {
        List<Integer> output = new ArrayList<>();
        if (rootNode == null) return output;

        Stack<NodeVisitPair> travelStack = new Stack<>();
        travelStack.push(new NodeVisitPair(rootNode, false));

        while (!travelStack.isEmpty()) {
            NodeVisitPair current = travelStack.pop();

            if (current.visited) {
                output.add(current.node.val);
            } else {
                current.visited = true;
                travelStack.push(current);

                List<Node> kids = current.node.children;
                for (int i = kids.size() - 1; i >= 0; i--) {
                    travelStack.push(new NodeVisitPair(kids.get(i), false));
                }
            }
        }

        return output;
    }

    private class NodeVisitPair {

        Node node;
        boolean visited;

        NodeVisitPair(Node node, boolean visited) {
            this.node = node;
            this.visited = visited;
        }
    }
}

This Java solution implements a postorder traversal on an N-ary tree where each node can have multiple children. The method postorderTraversal returns a list of integer values following the postorder traversal pattern using an iterative approach instead of recursion, which helps in managing memory more efficiently with deeper or more complex tree structures.

• Utilize a helper class NodeVisitPair to manage the node and its visitation status during traversal.
• Initiate an empty list output to store the postorder traversal result.
• Use a stack travelStack to manage the nodes during traversal, where unwinding only happens once all children of the node are processed.
• Iterate through the stack until it is empty:
  • Pop the top NodeVisitPair element from the stack.
  • Check if the node in this pair has been visited.
  • If it has been visited, add the node's value to the output list.
  • If it hasn't been visited, mark it as visited, push it back onto the stack, and then push all its children onto the stack in reverse order (to maintain the correct order for postorder traversal when they are popped off the stack).

Ensure that the starting node (root node) is non-null to avoid processing an empty tree. This method effectively handles the postorder traversal of nodes and correctly assembles the traversal order into the output list which is finally returned.

class Solution:
    def traversal_postorder(self, root: "Node") -> List[int]:
        output = []
        if root is None:
            return output
        
        stack = [(root, False)]
        
        while stack:
            node, visited = stack.pop()
            
            if visited:
                output.append(node.val)
            else:
                stack.append((node, True))
                for child in reversed(node.children):
                    stack.append((child, False))
        
        return output

This summary explains the solution implemented in Python for postorder traversal of an N-ary Tree, which involves visiting each node's children before the node itself.

Understand that the primary data structure used is a stack to manage the traversal order dynamically. The function traversal_postorder first checks if the root node is None, immediately returning an empty list if true since no nodes exist to traverse.

Initialize an empty list named output to store the traversal result and a stack initialized with a tuple containing the root node and a Boolean flag set to False. The Boolean flag in each tuple indicates whether the node has been visited.

The iteration over the stack continues until it is empty, where in each iteration:

• The node and its visited status are retrieved by popping from the stack.
• If the node is marked as visited (True), append the node's value to the output list.
• If the node is not visited (False):
  • Push the node back onto the stack with the visited status set to True.
  • Push all children of the node onto the stack in reversed order (ensuring left-to-right child processing in postorder) with their visited status set to False.

The process ensures each node is revisited only after all its children are processed, adhering to postorder traversal's requirements. Finally, the output list, containing the values of the nodes in postorder, is returned.