有点感动，这题在AS里写好之后复制过去一次AC了。 这种BFS的套路跟这题的一样，用queue，维护curNum和nextNum；我跟code ganker学会了这个套路，按照覃超的标准来说这代码是有点长的，但是现在已经形成记忆了，就不去修改了。思路还蛮清晰的。

public List
    
     
      > levelOrderBottom(TreeNode root) {        List
      
       
        > res = new ArrayList<>();        if (root == null) return res;        LinkedList
        
          queue = new LinkedList<>();        List
         
           cell = new ArrayList<>(); queue.add(root); int curNum = 1; int nextNum = 0; while (!queue.isEmpty()) { TreeNode node = queue.poll(); cell.add(node.val); curNum--; if (node.left != null) { queue.offer(node.left); nextNum++; } if (node.right != null) { queue.offer(node.right); nextNum++; } if (curNum == 0) { res.add(0, cell); curNum = nextNum; nextNum = 0; cell = new ArrayList<>(); } } return res; }复制代码
         
        
       
      
     
    

另

贴一下leetcode里面的人的简洁写法：https://discuss.leetcode.com/topic/7651/my-dfs-and-bfs-java-solution/2

public class Solution {    public List
    
     
      > levelOrderBottom(TreeNode root) {        Queue
      
        queue = new LinkedList
       
        ();        List
        
         
          > wrapList = new LinkedList
          
           
            >(); if(root == null) return wrapList; queue.offer(root); while(!queue.isEmpty()){ int levelNum = queue.size(); List
            
              subList = new LinkedList
             
              (); for(int i=0; i
             
            
           
          
         
        
       
      
     
    

递归（我没看）:

public class Solution {        public List
    
     
      > levelOrderBottom(TreeNode root) {            List
      
       
        > wrapList = new LinkedList
        
         
          >(); levelMaker(wrapList, root, 0); return wrapList; } public void levelMaker(List
          
           
            > list, TreeNode root, int level) { if(root == null) return; if(level >= list.size()) { list.add(0, new LinkedList
            
             ()); } levelMaker(list, root.left, level+1); levelMaker(list, root.right, level+1); list.get(list.size()-level-1).add(root.val); } }复制代码