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 that for each level of the tree, we alternate the direction of traversal. The first level is traversed from left to right, the second from right to left, the third from left to right, and so on.
This type of traversal is significant in scenarios where the order of processing nodes alternates, such as in certain search algorithms or visual representations of tree structures.
Potential pitfalls include not correctly alternating the traversal direction or not handling null nodes properly.
To solve this problem, we can use a breadth-first search (BFS) approach with a queue to keep track of nodes at each level. We will also use a flag to alternate the direction of traversal for each level.
1. **Naive Solution**: A naive solution might involve using a simple BFS without alternating directions, but this would not meet the problem requirements.
2. **Optimized Solution**: The optimized solution involves using a deque (double-ended queue) to efficiently append nodes from both ends based on the current traversal direction.
1. Initialize a queue with the root node and a flag for direction (left-to-right initially).
2. While the queue is not empty, process each level:
3. Append the current level list to the result list.
from collections import deque
# 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 = deque([root])
left_to_right = True
while node_queue:
level_size = len(node_queue)
current_level = deque()
for _ in range(level_size):
node = node_queue.popleft()
if left_to_right:
current_level.append(node.val)
else:
current_level.appendleft(node.val)
if node.left:
node_queue.append(node.left)
if node.right:
node_queue.append(node.right)
results.append(list(current_level))
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 we visit each node exactly once. The space complexity is also O(n) due to the storage required for the queue and the result list.
1. An empty tree (root is null) should return an empty list.
2. A tree with only one node should return a list with one sublist containing that node's value.
3. Trees with varying depths and structures should be tested to ensure the zigzag pattern is maintained.
To test the solution comprehensively, consider the following test cases:
Using a testing framework like `unittest` in Python can help automate and validate these test cases.
1. Break down the problem into smaller parts and understand each requirement.
2. Use diagrams to visualize the tree structure and the traversal process.
3. Practice similar tree traversal problems to strengthen your understanding.
Understanding and solving the zigzag level order traversal problem helps improve your grasp of tree traversal techniques and alternating patterns. Practice and explore further to enhance your problem-solving skills.