Binary Tree Zigzag Level Order Traversal

 

1. Clarify the problem: Before diving into solving the problem, let's clarify the requirements:

  • The input is a binary tree.
  • We need to traverse the tree in a zigzag pattern, starting from the root level.
  • For each level, we alternate between going from left to right and right to left.
  • The output should be a list of lists, where each inner list represents the nodes at a particular level in the tree.

2. Analyze the problem: To solve this problem, we can perform a level order traversal of the binary tree and modify the order of traversing the nodes at each level based on whether the level is odd or even. We can use a queue to perform the level order traversal, and a flag to keep track of the current level's order (left to right or right to left).

3. Design an algorithm: Here is the algorithm to solve the problem:

  1. Create an empty result list to store the zigzag level order traversal.
  2. If the root is null, return the result list.
  3. Create an empty queue and enqueue the root node.
  4. Create a flag variable to keep track of the level's order, starting with left to right (true).
  5. While the queue is not empty:
    • Get the size of the queue (to process all nodes at the current level).
    • Create a new level list to store the nodes at the current level.
    • Iterate through the nodes at the current level:
      • Dequeue a node from the front of the queue.
      • Depending on the level's order, either append the node to the level list (if left to right) or insert it at the beginning (if right to left).
      • Enqueue the left and right child nodes of the dequeued node (if they exist).
    • Append the level list to the result list.
    • Toggle the level's order flag.
  6. Return the result list.

4. Explain your approach: The approach uses a level order traversal technique with a slight modification to the order of nodes at each level based on whether the level is odd or even. By toggling the level's order flag, we can alternate between left to right and right to left at each level.

5. 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 zigzagLevelOrder(root):
    if not root:
        return []

    result = []
    queue = [root]
    left_to_right = True

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

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

            if left_to_right:
                level_nodes.append(node.val)
            else:
                level_nodes.insert(0, node.val)

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

        result.append(level_nodes)
        left_to_right = not left_to_right

    return result

6. Test your code: Let's test the code with the following test case:

python
# Example binary tree:
#        3
#       / \
#      9  20
#         / \
#        15  7

root = TreeNode(3)
root.left = TreeNode(9)
root.right = TreeNode(20)
root.right.left = TreeNode(15)
root.right.right = TreeNode(7)

print(zigzagLevelOrder(root))

The expected output is:

[[3], [20, 9], [15, 7]]

7. Optimize if necessary: The provided solution has a time complexity of O(N), where N is the number of nodes in the binary tree, since we need to traverse all the nodes once. The space complexity is also O(N), as in the worst case, the queue can hold all the nodes at a particular level.

8. Handle error cases: The code handles the case when the root is None. If the tree contains invalid nodes or values, the code does not explicitly handle such cases. Additional error handling can be implemented depending on the specific requirements.

9. Discuss complexity analysis:

  • Time complexity: The time complexity of the solution is O(N), where N is the number of nodes in the binary tree. We visit each node once during the level order traversal.
  • Space complexity: The space complexity is O(N) as well. In the worst case, the queue can hold all the nodes at a particular level. Additionally, the result list will contain N elements since each node is visited once.
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