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  1. Prepare
  2. Tutorials
  3. 30 Days of Code
  4. Day 23: BST Level-Order Traversal
  5. Discussions

Day 23: BST Level-Order Traversal

  • conorvoss7
     + 0 comments

    Since we can't use Python's collections.deque and the O(1) popleft function, it's actually more effecient to just maintain a pointer to your current location in the list (this avoids the O(N) cost of each pop(0) call).

    Python 3:

        def levelOrder(self,root):
        q = []
        if root is not None:
            q.append(root)
        i = 0
        while i < len(q):
            n = q[i]
            if n.left is not None:
                q.append(n.left)
            if n.right is not None:
                q.append(n.right)
            i += 1
        print(" ".join(str(x.data) for x in q))