Binary Tree Level Order Traversal II

Question

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).

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]
]

Answer

solution：

# 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 levelOrderBottom(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root:
            return []
        ans , level = [],[root]
        while level:
            ans.append([node.val for node in level]) #cause append just add a new element, if list , the list will be an element.
            temp = []
            for node in level:
                temp.extend([node.left,node.right])
            level = [leaf for leaf in temp if leaf]
        return ans[::-1]

Knowledge：

1. 这道题目，我偷了个懒，将上一题的列表进行逆向索引。

2. 切片操作符在python中的原型是 [start:stop:step] 即:[开始索引:结束索引:步长值]，步长值为正时表示从左向右取，如果为负，则表示从右向左取。所以[::-1]省略起始索引、终止索引，步长值为-1，表示反向获取所有值。