剑60—把二叉树打印成多行(按层)

题目描述

从上到下按层打印二叉树,同一层结点从左至右输出。每一层输出一行。

思想

二叉树打印,队列思想
1、新建当前所在层节点的队列queue,根节点入队列

2、如果队列queue不为空

2.1新建缓冲队列buff,存放当前节点的孩子节点
新建集合al,存放当前层的节点
2.2 如果队列queue不为空,

  1. poll出栈顶元素即为cur,将当前元素值加入集合al
    2.2.1 若cur的左孩子不为空,将左孩子加入缓冲队列buff
    2.2.2 若cur的右孩子不为空,将右孩子加入缓冲队列buff

2.3 否则当前队列queue为空,说明当前层节点遍历完,则将al加入res,
并将buff赋值给queue,成为父节点继续循环

3、返回res

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;

/*
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<>();
        if (pRoot == null)
            return result;

        Queue<TreeNode> queue = new LinkedList<>();//当前所在层的节点的队列
        queue.offer(pRoot);

        while (!queue.isEmpty()) //遍历每一层
        {
            Queue<TreeNode> buff = new LinkedList<>();//新建缓冲队列:当前节点的孩子节点的队列
            ArrayList<Integer> al = new ArrayList<>();//新建集合al,存放当前层节点

            while (!queue.isEmpty()) //遍历当前层
            {
                TreeNode cur = queue.poll();
                al.add(cur.val);
                if (cur.left != null) 
                    buff.offer(cur.left);
                if (cur.right != null)
                    buff.offer(cur.right);
            }
            result.add(al);
            queue = buff;//转向遍历下一层
        }

        return result;
    }
 /*   ArrayList<ArrayList<Integer>> Print(TreeNode pRoot) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<>();
        if (pRoot == null) return result;
        int curRow=1;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(pRoot);
        while (!queue.isEmpty()) {
            int nextRow=0;
            ArrayList<Integer> al = new ArrayList<>();
            for (int i = 0; i < curRow; i++) {
                TreeNode cur = queue.poll();
                al.add(cur.val);
                if (cur.left != null) {
                    queue.offer(cur.left);
                    nextRow++;
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                    nextRow++;
                }
            }

            result.add(al);
            curRow = nextRow;
        }
        return result;
    }*/
}

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转载自blog.csdn.net/hefenglian/article/details/80114758