Question

Print a binary tree layer-by-layer from top to bottom, and from left to right for each layer.

Solution

Yes, it’s a simple task. We can use breadth-first search, and which means we need a queue, see reference.

levelorder(root)
    q = empty queue
    q.enqueue(root)
    while not q.empty do
        node := q.dequeue()
        visit(node)
        if node.left ≠ null then
            q.enqueue(node.left)
        if node.right ≠ null then
            q.enqueue(node.right)

Example

# node structure
class bst_node:
    value = None
    left = None
    right = None
    def __init__(self, value, left=None, right=None):
        self.value = value
        self.left = left
        self.right = right

# what to do with a node.
def visit(node):
    print node.value,

# bfs traversal
def breadth_first_traversal(root):
    queue = [root]
    while len(queue) > 0:
        node = queue.pop(0) # get the head item
        visit(node)
        if node.left: queue.append(node.left)
        if node.right: queue.append(node.right)

# a binary tree and display how was it before reflection
node_05 = bst_node(5)
node_07 = bst_node(7)
node_09 = bst_node(9)
node_11 = bst_node(11)
node_06 = bst_node(6,  node_05, node_07)
node_10 = bst_node(10, node_09, node_11)
node_08 = bst_node(8,  node_06, node_10)

# answer should be: 8 6 10 5 7 9 11
root = node_08
breadth_first_traversal(root)

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