Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
Example:
Input: root = [3, 9, 20, null, null, 15, 7]
3
/ \
9 20
/ \
15 7
Output:
[
[3],
[20,9],
[15,7]
]
The core challenge of this problem is to traverse the binary tree in a zigzag manner. This means alternating the direction of traversal at each level of the tree. This type of traversal is significant in scenarios where hierarchical data needs to be processed in a non-linear fashion.
Common applications include visualizing hierarchical data structures, such as organizational charts or file systems, where a zigzag pattern can provide a more intuitive view.
Potential pitfalls include handling null nodes and ensuring the correct order of nodes at each level.
To solve this problem, we can use a breadth-first search (BFS) approach with a queue to traverse the tree level by level. We will use a flag to alternate the direction of traversal at each level.
Initial naive solution might involve using a simple BFS without considering the zigzag pattern, but this would not meet the problem requirements.
Optimized solutions involve using a deque to efficiently append nodes from both ends and a flag to track the direction of traversal.
import collections
# Definition for a binary tree node.
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 []
results = []
node_queue = collections.deque([root])
left_to_right = True
while node_queue:
level_size = len(node_queue)
level_nodes = collections.deque()
for _ in range(level_size):
node = node_queue.popleft()
if left_to_right:
level_nodes.append(node.val)
else:
level_nodes.appendleft(node.val)
if node.left:
node_queue.append(node.left)
if node.right:
node_queue.append(node.right)
results.append(list(level_nodes))
left_to_right = not left_to_right
return results
# Example usage:
# Construct the binary tree from the example
root = TreeNode(3)
root.left = TreeNode(9)
root.right = TreeNode(20)
root.right.left = TreeNode(15)
root.right.right = TreeNode(7)
print(zigzagLevelOrder(root)) # Output: [[3], [20, 9], [15, 7]]
The time complexity of this approach is O(N), where N is the number of nodes in the tree. This is because each node is processed exactly once.
The space complexity is also O(N) due to the storage required for the queue and the result list.
Potential edge cases include:
These cases are handled by the algorithm as it processes each node individually and does not assume a balanced tree.
To test the solution comprehensively, consider the following test cases:
root = None
root = TreeNode(1)
root = [3, 9, 20, null, null, 15, 7]
root = [1, 2, 3, 4, null, null, 5]
Use a testing framework like unittest
or pytest
to automate the testing process.
When approaching such problems, consider the following tips:
Understanding and solving the zigzag tree traversal problem enhances your knowledge of tree data structures and traversal algorithms. It also improves your problem-solving skills by challenging you to think about different traversal patterns and their implementations.
Practice regularly and explore further to deepen your understanding and proficiency in solving such problems.