Binary Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

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

URL: https://leetcode.com/problems/binary-tree-level-order-traversal-ii/

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

import Queue
class Solution:
    # @param {TreeNode} root
    # @return {integer[][]}
    def levelOrderBottom(self, root):
        if root == None:
            return []
        else:
            q = Queue.Queue()
            q.put(root)
            q.put("#")
            levelOrderTraversal = []
            level = []
            stack = []

            while q.empty() == False:
                node = q.get()
                if node == "#":
                    if q.empty() == False:
                        q.put("#")
                    stack.append(level)
                    level = []
                else:
                    level.append(node.val)
                    if node.left:
                        q.put(node.left)
                    if node.right:
                        q.put(node.right)

            while stack:
                levelOrderTraversal.append(stack.pop())

            return levelOrderTraversal