Sword finger offer 32 - ii Print binary tree II from top to bottom

Title Description

Problem analysis

This problem uses the sequence traversal of breadth first traversal , Use iterative method to traverse ( According to the characteristics of sequence traversal, queues are used for storage )
The difference between this question and the above one is that the output is a two-dimensional linked list , That is, use the result of the previous method , Introduce a layer while Loop through and store the value of each layer

Code instance

class Solution {
    
    public List<List<Integer>> levelOrder(TreeNode root) {
    
        List<List<Integer>> list = new ArrayList<>();
        if(root == null){
    
        	return null; 
        }
        Queue<TreeNode> que = new LinkedList<TreeNode>();
        que.offer(root);
        while(!que.isEmpty()){
    
        	List<Integer> itemlist = new ArrayList<>();
        	TreeNode tempnode = queue.poll();
        	int len = que.size();
        	while(len > 0){
    
        		if(tempnode.left != null){
    
        			que.offer(tempnode.left);
        		}
        		if(tempnode.right != null){
    
        			que.offer(tempnode.right);
        		}
        		len--;
        	}
        	list.add(itemlist);
        }
        return list;
    }
}