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;
    }
}