[leetcode]Binary Tree Level Order Traversal II

问题描写叙述：

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.

代码：

public List<List<Integer>> levelOrderBottom(TreeNode root) {  //java
		if(root == null)
			return new ArrayList<>();
			
		Stack<TreeNode> stack = new Stack<TreeNode>();
		stack.add(root);
		List<TreeNode> levelList = new ArrayList<TreeNode>();
		
		
		List<List<Integer>> result = new ArrayList<List<Integer>>();
		List<List<Integer>> invert_result = new ArrayList<List<Integer>>();
		List<Integer> tmp = new ArrayList<Integer>();
		
		while(!stack.isEmpty()|| !levelList.isEmpty()){
			if(stack.isEmpty()){
				for(int i=levelList.size()-1; i>=0; i--)
					stack.push(levelList.get(i));
				levelList.clear();
				invert_result.add(tmp);
				tmp = new ArrayList<Integer>();
			}
			
			while(!stack.isEmpty()){
				TreeNode node = stack.pop();
				tmp.add(node.val);
				
				if(node.left !=null)
					levelList.add(node.left);
				if(node.right !=null)
					levelList.add(node.right);
			}
		}
		invert_result.add(tmp);
		
		//invert result 
		for(int i=invert_result.size()-1; i>=0; i--)
			result.add(invert_result.get(i));
		
		return result;
        
    }