原题地址:http://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/
题意:
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,#,#,15,7}
,
3 / \ 9 20 / \ 15 7
return its bottom-up level order traversal as:
[ [15,7] [9,20], [3], ]
confused what "{1,#,2,3}"
means? > read more on how binary tree is serialized on OJ.
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
1 / \ 2 3 / 4 \ 5The above binary tree is serialized as
"{1,2,3,#,#,4,#,#,5}"
.
# 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 preorder(self, root, level, res): if root: if len(res) < level+1: res.append([]) res[level].append(root.val) self.preorder(root.left, level+1, res) self.preorder(root.right, level+1, res) def levelOrderBottom(self, root): res=[] self.preorder(root, 0, res) res.reverse() return res