[leetcode]Binary Tree Level Order Traversal II

问题描述：

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) {  //javaif(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;            }