把二叉树打印成多行

利用层次遍历的算法，设置变量count与last，用count来保存当前层已经访问的结点数，last用来保存当前层的最大结点数。当count=last时，表示该层访问结束。

 

import java.util.*;


/*
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/
public class Solution {
    ArrayList<ArrayList<Integer> > Print(TreeNode pRoot) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
        if(pRoot == null)
            return result;
        TreeNode current = pRoot;
        queue.offer(current);
        int count; //用来记录当前节点
        int last;    //用来记录每一层节点数
        while(!queue.isEmpty()){
            count = 0;
            last = queue.size();
            ArrayList<Integer> list = new ArrayList<Integer>();
            //打印一层
            while(count < last){
                current = queue.pop();
                list.add(current.val);
                count++;
                //这里直接往queue里添加下一层节点，为什么不影响当前while循环里的queue.pop()？？？
                if(current.left != null){
                    queue.offer(current.left);
                }
                if(current.right != null){
                    queue.offer(current.right);
                }
            }
            result.add(list);
        }
        return result;
    }
    
}

while循环里有一点不懂