Java for LeetCode 107 Binary Tree 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,#,#,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;
    }