Python Binary Search Tree Level Order | David Capella
View on GitHub

David Capella

Data Engineer

Python Binary Search Tree Level Order


A level-order traversal, also known as a breadth-first search, visits each level of a tree’s nodes from left to right, top to bottom. You are given a pointer, root, pointing to the root of a binary search tree. Complete the levelOrder function provided in your editor so that it prints the level-order traversal of the binary search tree. Hint: You’ll find a queue helpful in completing this challenge.


Honestly this one had me stumped for awhile. This is what I started off with. My final solution came from the second image; the code all the way to the right.

bst-level-order bst-level-order2

It was actually very simple but I was thinking about it too much. At first I instantly wanted to do recursion but quickly found out that, that was the wrong answer simply because I needed to print, not return anything.

That means either a for loop or a while loop. I had trouble at first thinking about how to go about it. Until I realized I was trying to do too much at one time. I wanted to go through each node and child node while printing out the data for each one.

Once I figured out that I need to first add all the nodes in the right order then print them out, it became easier.

Still, I tried a for loop inside a while loop (never can start off easy, huh?), realized that was wrong, and switched to my final solution of a simple while loop, while (no pun intended) keeping track of the index. Obviously should not use a for loop when I am adding to the list that I am trying to loop over.

I love ‘try and except’ but I did not think about them at first during this. Probably still dazed from trying to do extra work.

Here we go.


def levelOrder(self,root):
  #Write your code here
  # Decalre all the nodes (root for now) and index starting
  nodes = [root]
  i = 0
  # Time to loop, normally this is not a good idea but I am using a try/except
  while True:
    # This try/except will alow us to break the while loop if no more nodes to add
    # because it will be index out of bounds
      # Load the first node since no for loop    
      node = nodes[i]
      # Load up the nodes with the children if any
      if node.left: nodes.append(node.left)
      if node.right: nodes.append(node.right)
      # Make sure to keep count of where we are
      i += 1
  # Out of the while loop time to do the actual problem and print each data
  for n in nodes:
    print(, end=" ")


Definitely made it harder than it needed to be. But that is a lesson in itself. Multiple lessons learned, check.