题目描述

• 按层次，那么先按序读完本层值，并且存入队列中，再根据队列走下一层的值即可
  

代码 & 思路

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public List<List<Integer>> levelOrder(TreeNode root) {List<List<Integer>> ans = new ArrayList<>();if(root != null) {LinkedList<TreeNode> queue = new LinkedList<>();queue.offer(root);findAns(ans, 1, queue);}return ans;}// 一次找一层void findAns(List<List<Integer>> ans, int nodeNums, LinkedList<TreeNode> queue){if(queue.size() == 0) {return;}int nowNum = 0;List<Integer> nowList = new ArrayList<>();for(int i = 0; i < nodeNums; i++) {TreeNode nowNode = queue.removeFirst();nowList.add(nowNode.val);// 处理左右子结点if(nowNode.left != null){queue.offer(nowNode.left);nowNum++;}if(nowNode.right != null){queue.offer(nowNode.right);nowNum++;}}ans.add(nowList);findAns(ans,nowNum,queue);}
}