乍一看我一脸懵逼，不知如何下手。但仔细想一下，方法还是有点意思。前面几道题都是树，树的遍历是很常见的题目。这道题采取先序遍历（当然，中序和后续遍历也是可以的，且只用改变一行代码），每向下遍历一层，就新建一个列表并编号，这样，每一层就有一个对应的集合，遍历到相应层的元素，放到对应的集合中去。

/**
 * 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>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        levelHelper(res, root, 0);
        return res;
    }
    
    public void levelHelper(List<List<Integer>> res, TreeNode root, int height) {
        if (root == null) return;
        if (height >= res.size()) {
            res.add(new LinkedList<Integer>());
        }
        res.get(height).add(root.val);//It’s position could be changed
        levelHelper(res, root.left, height+1);
        levelHelper(res, root.right, height+1);
    }
}

/**
 * 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>> zigzagLevelOrder(TreeNode root) {
	return zigzaLevelHeapler(root, new ArrayList<List<Integer>>(), 0);
}

public List<List<Integer>> zigzaLevelHeapler(TreeNode root, List<List<Integer>> ans, int level) {
	if (root == null) {
		return ans;
	}
	if (level  > ans.size() -1)
		ans.add(level, new ArrayList<Integer>());

	List<Integer> levelList = ans.get(level);

	if (level % 2 == 0) {
		levelList.add(root.val);
	} else {
		levelList.add(0, root.val);
	}
	zigzaLevelHeapler(root.left, ans, level + 1);
	zigzaLevelHeapler(root.right, ans, level + 1);
	return ans;
}
}

**
 * 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>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        
        if(root==null)
            return res;
        
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        int count =1;
        while(!queue.isEmpty()){
            int size = queue.size();
            List<Integer> list = new ArrayList<Integer>();
            for(int i=0; i<size; i++){
                TreeNode node = queue.remove();
                list.add(node.val);
                if(node.left!=null)
                    queue.add(node.left);
                if(node.right!=null)
                    queue.add(node.right);
            }
            
            if(count%2==0){
                Collections.reverse(list);    
            }
            count++;
            res.add(list);
        }
        return res;
    }
}

BFS solution：

/**
 * 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>> wrapList = new LinkedList<List<Integer>>();
            levelMaker(wrapList, root, 0);
            return wrapList;
        }
        
        public void levelMaker(List<List<Integer>> list, TreeNode root, int level) {
            if(root == null) return;
            if(level >= list.size()) {
                list.add(0, new LinkedList<Integer>());
            }
            list.get(list.size()-level-1).add(root.val);
            levelMaker(list, root.left, level+1);
            levelMaker(list, root.right, level+1);
            
        }
}

DFS solution：

public class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        List<List<Integer>> wrapList = new LinkedList<List<Integer>>();
        
        if(root == null) return wrapList;
        
        queue.offer(root);
        while(!queue.isEmpty()){
            int levelNum = queue.size();
            List<Integer> subList = new LinkedList<Integer>();
            for(int i=0; i<levelNum; i++) {
                if(queue.peek().left != null) queue.offer(queue.peek().left);
                if(queue.peek().right != null) queue.offer(queue.peek().right);
                subList.add(queue.poll().val);
            }
            wrapList.add(0, subList);
        }
        return wrapList;
    }
}