For example:
Given binary tree
    3
   / \
  9  20
    /  \
   15   7
return its  level order traversal as:
[[15,7],[9,20],[3] ]

 Approach:
1.Add root to queue 
2.pop item in queue and add left and right nodes of that item to queue.now add value of popped to a temporary array
3.repeat the above step len(queue) no of times
4.Add that temporary to result array
5.if size of queue !=0 go to step 2
6.return reversed result array
Python Code:
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def levelOrderBottom(self, root):
        if not root :
            return []
        queue = [] 
        result = []
        queue.append(root)
        while len(queue)>0:
            temp = []
            for i in range(len(queue)):
                node = queue.pop(0)
                temp.append(node.val)
                if node.left!=None: queue.append(node.left)
                if node.right!=None: queue.append(node.right)
            result.insert(0,temp)

        return result

 Try Quoeston Here