Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level).
Example 1:
Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[15,7]]
Example 2:
Input: root = [1]
Output: [[1]]
Example 3:
Input: root = []
Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 2000]. -
-1000 <= Node.val <= 1000
SOLUTION:
# 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]]:
if not root:
return []
queue = [(root, 0)]
levels = {}
level = 0
while len(queue) > 0:
curr, level = queue.pop(0)
levels[level] = levels.get(level, []) + [curr.val]
if curr.left:
queue.append((curr.left, level + 1))
if curr.right:
queue.append((curr.right, level + 1))
return [levels[l] for l in range(level + 1)]
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