# Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
#
# For example:
# Given binary tree [3,9,20,null,null,15,7],
# 3
# / \
#     9  20
# / \
#     15   7
# return its level order traversal as:
# [
#     [3],
#     [9,20],
#     [15,7]
# ]

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Solution(object):
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root:
            return []
        queue = [root]
        result = []
        while queue:
            temp = [x.val for x in queue]
            new_queue = []
            for node in queue:
                if node.left:
                    new_queue.append(node.left)
                if node.right:
                    new_queue.append(node.right)
            queue = new_queue
            result.append(temp)
        return result

    def levelOrderRecursive(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        self.levels = {}
        self.traverse(root, 0)
        return [self.levels[key] for key in sorted(self.levels.keys())]

    def traverse(self, r, c):
        if r:
            self.levels[c] = self.levels.get(c, []) + [r.val]
            self.traverse(r.left, c+1)
            self.traverse(r.right, c+1)

# Note:
# Iterative solution using BFS and intuitive recursive solution
