Performing Breadth First Search recursively

Question

Let\'s say you wanted to implement a breadth-first search of a binary tree recursively. How would you go about it?

Is it possible using only the call-stack

Answer 1

If you use an array to back the binary tree, you can determine the next node algebraically. if i is a node, then its children can be found at 2i + 1 (for the left node) and 2i + 2 (for the right node). A node's next neighbor is given by i + 1, unless i is a power of 2

Here's pseudocode for a very naive implementation of breadth first search on an array backed binary search tree. This assumes a fixed size array and therefore a fixed depth tree. It will look at parentless nodes, and could create an unmanageably large stack.

bintree-bfs(bintree, elt, i)
    if (i == LENGTH)
        return false

    else if (bintree[i] == elt)
        return true

    else 
        return bintree-bfs(bintree, elt, i+1)        

Answer 2

I couldn't find a way to do it completely recursive (without any auxiliary data-structure). But if the queue Q is passed by reference, then you can have the following silly tail recursive function:

BFS(Q)
{
  if (|Q| > 0)
     v <- Dequeue(Q)
     Traverse(v)
     foreach w in children(v)
        Enqueue(Q, w)    

     BFS(Q)
}

Answer 3

Here is a BFS recursive traversal Python implementation, working for a graph with no cycle.

def bfs_recursive(level):
    '''
     @params level: List<Node> containing the node for a specific level.
    '''
    next_level = []
    for node in level:
        print(node.value)
        for child_node in node.adjency_list:
            next_level.append(child_node)
    if len(next_level) != 0:
        bfs_recursive(next_level)


class Node:
    def __init__(self, value):
        self.value = value
        self.adjency_list = []