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],

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5
  3
 / \
9  20
  /  \
 15   7

return its bottom-up level order traversal as:

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[
  [15,7],
  [9,20],
  [3]
]

思路

102题也是一道层次遍历的题。在102题中，题目要求是采用自顶向下的输出方式；而这一题采用了自底向上的输出方式。

由于两道题思路相同，都是在队列访问的时候加入空节点用于分割，就不再赘述了。代码后面附上参考链接。

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# 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]]
        """
        q = []
        res = []
        layer = []
        if (root):
            q.append(root)
            q.append(None)
        else:
            return q
        while (len(q) > 0):
            node = q[0]
            layer.append(q[0])
            del q[0]
            if (node is not None):
                if (node.left is not None):
                    q.append(node.left)
                if (node.right is not None):
                    q.append(node.right)
            else:
                layerval = []
                if(len(q) > 0):
                    q.append(None)
                    for i in layer:
                        if (i):
                            layerval.append(i.val)
                    res.append(layerval)
                    layer = []
        lastlayer = []
        for i in layer:
            if (i):
                lastlayer.append(i.val)
        res.append(lastlayer)
        
        res.reverse()
        return res

参考链接：http://www.cnblogs.com/miloyip/archive/2010/05/12/binary_tree_traversal.html