Traversing a n-ary tree without using recurrsion

Question

How can I traverse an n-ary tree without using recursion?

Recursive way:

traverse(Node node)
{
    if(node == null)
        return;

            


        

Answer 1

No language given, so in pseudo-pseudocode:

traverse(Node node)
{
  List<Node> nodes = [node];

  while (nodes.notEmpty) {
    Node n = nodes.shift();

    for (Node child in n.getChildren()) {
      nodes.add(child);
    }

    // do stuff with n, maybe
  }
}

Note that this is a breadth-first traversal as opposed to the depth-first traversal given in the question. You should be able to do a depth-first traversal by poping the last item off the nodes list instead of shifting the first one.

Answer 2

What you are doing is essentially a DFS of the tree. You can eliminate recursion by using a stack:

traverse(Node node) {
    if (node==NULL)
        return;

    stack<Node> stk;
    stk.push(node);

    while (!stk.empty()) {
        Node top = stk.pop();
        for (Node child in top.getChildren()) {
            stk.push(child);
        }
        process(top);
    }
}

If you want a BFS use a queue:

traverse(Node node) {
    if (node==NULL)
        return;

    queue<Node> que;
    que.addRear(node);

    while (!que.empty()) {
        Node front = que.deleteFront();
        for (Node child in front.getChildren()) {
            que.addRear(child);
        }
        process(front);
    }
}

In case you want some other way to traverse, you'll have to follow the same approach albeit with a different data structure to store the node. Maybe a priority queue (if you want to evaluate a function at each node and then process nodes based on that value).

Answer 3

You can do this without recursion and without a stack. But you need to add two extra pointers to the node:

1. The parent node. So you can come back to the parent if you are finished.

2. The current child node so you know which one to take next.

   • For each node, you handle all the kids.
   • If a kid is handled, you check if there is a next kid and handle that (updating the current).
   • If all kids are handled, go back to the parent.
   • If the node is NULL, quit.

With pseudocode this looks like:

traverse(Node node) {
  while (node) {
    if (node->current <= MAX_CHILD) {
      Node prev = node;
      if (node->child[node->current]) {
        node = node->child[node->current];
      }
      prev->current++;
    } else {
      // Do your thing with the node.
      node->current = 0; // Reset counter for next traversal.
      node = node->parent;
    }
  }
}