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:
3
/ \
9 20
/ \
15 7
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
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# 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]]:
def traverse(root, level, result):
if not root:
return
if level > len(result)-1:
result.append([])
result[level].append(root.val)
traverse(root.left, level+1, result)
traverse(root.right, level+1, result)
result = []
traverse(root, 0, result)
return result
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