Binary Tree Level Order Traversal - Solution
Solutions and explanations
Video Explanation
# 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
class Solution:
def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
res = []
if not root:
return res
queue = collections.deque([root])
while queue:
cur_level_res = []
for _ in range(len(queue)):
node = queue.popleft()
# explore node
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
cur_level_res.append(node.val)
res.append(cur_level_res)
return res
Complexity Analysis
- Time Complexity:
O(n)time is required, as each node is visited. - Space Complexity:
O(n)space is used for queue, and for the result array (or list).
Here, n is the number of nodes in the binary tree.