摘要：
=null）queue.add（child.left）；如果（child.right！
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] ]
解题思路：
修改下Java for LeetCode 102 Binary Tree Level Order Traversal即可解决，JAVA实现如下：
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> list = new ArrayList<List<Integer>>();
if (root == null)
return list;
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
while (queue.size() != 0) {
List<Integer> alist = new ArrayList<Integer>();
for (TreeNode child : queue)
alist.add(child.val);
list.add(new ArrayList<Integer>(alist));
Queue<TreeNode> queue2=queue;
queue=new LinkedList<TreeNode>();
for(TreeNode child:queue2){
if (child.left != null)
queue.add(child.left);
if (child.right != null)
queue.add(child.right);
}
}
Collections.reverse(list);
return list;
}