Binary tree level order traversal

Time: O(N); Space: O(N); medium

Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

Example 1:

Input: {3,9,20,#,#,15,7}

  3
 / \
9  20
  /  \
 15   7

Output:

[
   [3],
   [9,20],
   [15,7]
]

[1]:

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

[2]:

class Solution1(object):
    def levelOrder(self, root: TreeNode):
        '''
        :type root: TreeNode
        :rtype: List[List[int]]
        '''
        if root is None:
            return []
        result, current = [], [root]
        while current:
            next_level, vals = [], []
            for node in current:
                vals.append(node.val)
                if node.left:
                    next_level.append(node.left)
                if node.right:
                    next_level.append(node.right)
            current = next_level
            result.append(vals)
        return result

[3]:

if __name__ == "__main__":

    root = TreeNode(3)
    root.left = TreeNode(9)
    root.right = TreeNode(20)
    root.right.left = TreeNode(15)
    root.right.right = TreeNode(7)

    assert Solution1().levelOrder(root) == [[3], [9, 20], [15, 7]]