107. Binary Tree Level Order Tra

对于这道BFS题目的两种不同解法（two queues or one queue）

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        if(root == null){
            return new ArrayList<>();
        }
        //注意初始化Queue的方法，一般实例为LinkedList
        Queue<TreeNode> nodeQueue = new LinkedList<TreeNode>();
        nodeQueue.add(root);
        List<Integer> tempList = new ArrayList<Integer>();
        List<List<Integer>> ans = new ArrayList<>();
        while(nodeQueue.size() > 0){
        //用size来track换行的时机
            int length = nodeQueue.size();
            tempList.clear();
            while(length > 0){
                TreeNode curHead = nodeQueue.poll();
                if(curHead.left != null){
                    nodeQueue.add(curHead.left);
                }
                if(curHead.right != null){
                    nodeQueue.add(curHead.right);
                }
                tempList.add(curHead.val);
                length--;
            }
        //必须新建一个与tempList相同的ArrayList添加入ans，否则后续会改变其内容
            ans.add(new ArrayList<Integer>(tempList));
        }
        //翻转ArrayList
        Collections.reverse(ans);
        return ans;
    }
}

用两个List的好处就是可以不需要track什么时候换行，递归一次就是一行结束

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> ans = new ArrayList<>();
        List<TreeNode> startList = new ArrayList<>();
        if(root == null){
            return new ArrayList<>();
        }
        startList.add(root);
        helper(startList, ans);
        Collections.reverse(ans);
        return ans;
    }
    
    public void helper(List<TreeNode> nodeList, List<List<Integer>> result){
        List<Integer> tempList = new ArrayList<Integer>();
        List<TreeNode> childList = new ArrayList<>();
        for(TreeNode ele: nodeList){
            tempList.add(ele.val);
            if(ele.left != null){
                childList.add(ele.left);
            }
            if(ele.right != null){
                childList.add(ele.right);
            }
        }
        result.add(tempList);
        //递归一次之后，如果子节点List不为空说明还有更深的节点，那么再递归一次
        if(childList.size() >= 1){
            helper(childList, result);
        }
        //反之，如若子节点List为空，则说明已经递归到了最深的一行，递归结束
    }
}