每日一题 102二叉树的层序遍历

题目

给你二叉树的根节点 root ，返回其节点值的 层序遍历 。 （即逐层地，从左到右访问所有节点）。

示例 1：

输入：root = [3,9,20,null,null,15,7]
输出：[[3],[9,20],[15,7]]
示例 2：

输入：root = [1]
输出：[[1]]
示例 3：

输入：root = []
输出：[]

提示：

树中节点数目在范围 [0, 2000] 内
-1000 <= Node.val <= 1000

题解

两个数组

/*** 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) {if (root == null) {return List.of();//建立一个空list}List<List<Integer>> ans = new ArrayList<>();List<TreeNode> cur = new ArrayList<>();cur.add(root);while (!cur.isEmpty()) {List<TreeNode> nxt = new ArrayList<>();List<Integer> vals = new ArrayList<>(cur.size());for(TreeNode node : cur) {vals.add(node.val);if (node.left != null) nxt.add(node.left);if (node.right != null) nxt.add(node.right);}cur = nxt;ans.add(vals);}return ans;}
}

队列

/*** 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) {if (root == null) {return List.of();}List<List<Integer>> ans = new ArrayList<>();Queue<TreeNode> q = new ArrayDeque<>();q.add(root);while (!q.isEmpty()) {int n = q.size();List<Integer> vals = new ArrayList<>(n);while (n-- > 0) {TreeNode node = q.poll();//删除队头的元素vals.add(node.val);if (node.left != null) q.add(node.left);if (node.right != null) q.add(node.right);}ans.add(vals);}return ans;}
}