Binary Tree Level Order Traversal

Clarify the problem:

• The problem requires us to perform a level order traversal of a binary tree and return the nodes at each level as a list of lists.
• Level order traversal visits the nodes at each level from left to right, starting from the root level.
• We need to implement a function that takes the root of the binary tree as input and returns the level order traversal as a list of lists.

Analyze the problem:

• Input: The root of a binary tree.
• Output: A list of lists containing the nodes at each level of the binary tree.
• Constraints:
  • The number of nodes in the binary tree is in the range [0, 1000].
  • The values of the nodes are unique.

Design an algorithm:

• We can use a breadth-first search (BFS) approach to perform the level order traversal.
• Create an empty list to store the result.
• Create an empty queue to store the nodes to be processed.
• Enqueue the root node into the queue.
• While the queue is not empty:
  • Create an empty list to store the nodes at the current level.
  • Get the current size of the queue (which represents the number of nodes at the current level).
  • Iterate through the nodes at the current level:
    • Dequeue a node from the queue.
    • Add the node's value to the list of nodes at the current level.
    • Enqueue the node's left and right children (if they exist) into the queue.
  • Append the list of nodes at the current level to the result list.
• Return the result list.

Explain your approach:

• We will implement a function called levelOrder to solve the problem.
• The function will take the root of the binary tree as input.
• We will create an empty list called result to store the level order traversal.
• We will create an empty queue called queue to store the nodes to be processed.
• We will enqueue the root node into the queue.
• While the queue is not empty, we will:
  • Create an empty list called level_nodes to store the nodes at the current level.
  • Get the current size of the queue.
  • Iterate i from 0 to the current queue size:
    • Dequeue a node from the queue.
    • Add the value of the dequeued node to level_nodes.
    • Enqueue the left and right children of the dequeued node (if they exist) into the queue.
  • Append level_nodes to result.
• Finally, we return result.

Write clean and readable code:

python

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def levelOrder(root):
    if not root:
        return []

    result = []
    queue = []
    queue.append(root)

    while queue:
        level_nodes = []
        level_size = len(queue)

        for _ in range(level_size):
            node = queue.pop(0)
            level_nodes.append(node.val)

            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)

        result.append(level_nodes)

    return result

Test your code:

• To test the code, we need to create a binary tree and verify that the level order traversal is correct.
• Let's create a binary tree with the following structure: 1 /
  2 3 / \
  4 5 6
• The level order traversal should be: [[1], [2, 3], [4, 5, 6]]
• We can use the following code to test the levelOrder function:

python

# Create the binary tree
root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(4)
root.left.right = TreeNode(5)
root.right.right = TreeNode(6)

# Perform level order traversal
result = levelOrder(root)

# Check if the level order traversal is correct
assert result == [[1], [2, 3], [4, 5, 6]]

We have tested the code with a sample binary tree and verified that the level order traversal is correct.

Optimize if necessary:

• The current solution has a time complexity of O(N), where N is the number of nodes in the binary tree.
• This is because we visit each node once during the level order traversal.
• The space complexity is also O(N) because in the worst case, the queue can hold all the nodes at the maximum level.

Handle error cases:

• The code handles the case when the root is None and returns an empty list in that case.
• The code assumes that the input is a valid binary tree.

Discuss complexity analysis:

• Let N be the number of nodes in the binary tree.
• The time complexity of the solution is O(N) because we visit each node once during the level order traversal.
• The space complexity is also O(N) because in the worst case, the queue can hold all the nodes at the maximum level.