# Time:  O(n)
# Space: O(n)
#
# 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,#,#,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:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution:
    # @param root, a tree node
    # @return a list of lists of integers
    def levelOrder(self, root):
        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

if __name__ == "__main__":
    root = TreeNode(3)
    root.left = TreeNode(9)
    root.right = TreeNode(20)
    root.right.left = TreeNode(15)
    root.right.right = TreeNode(7)
    result = Solution().levelOrder(root)
    print result