32.从上到下打印二叉树

参考链接

剑指offer_在线编程_ - 牛客网

LCR 149. 彩灯装饰记录 I - 力扣（LeetCode）

剑指 Offer 32 – I. 从上到下打印二叉树-跟着帅地玩转校招，刷爆各类算法题帅地玩Offer

个人尝试

/**
 * 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 int[] decorateRecord(TreeNode root) {
        if (root == null) return new int[0];

        List<Integer> list = new ArrayList<>();        
        Deque<TreeNode> queue = new ArrayDeque<>();
        // offer(E e) 队列已满时，返回 false，而不是抛异常
        // add(E e) 队列已满时，抛异常
        queue.addLast(root);
        while (!queue.isEmpty()) {
            // poll() 队列为空时返回 null，而不是抛异常
            // remove() 队列为空时抛异常
            TreeNode tmp = queue.removeFirst();
            list.add(tmp.val);
            if (tmp.left != null) queue.addLast(tmp.left);
            if (tmp.right != null) queue.addLast(tmp.right);
        }
        
        //这里可以用 Stream API
        //int[] result = list.stream().mapToInt(Integer::intValue).toArray();
        int[] result = new int[list.size()];
        for (int i = 0; i < result.length; i++) {
            result[i] = list.get(i);
        }
        return result;
    }
}

优秀题解

class Solution {
    public int[] decorateRecord(TreeNode root) {
        if(root == null) return new int[0];
        Queue<TreeNode> queue = new LinkedList<>(){{ add(root); }};
        ArrayList<Integer> ans = new ArrayList<>();
        while(!queue.isEmpty()) {
            TreeNode node = queue.poll();
            ans.add(node.val);
            if(node.left != null) queue.add(node.left);
            if(node.right != null) queue.add(node.right);
        }
        int[] res = new int[ans.size()];
        for(int i = 0; i < ans.size(); i++)
            res[i] = ans.get(i);
        return res;
    }
}

作者：Krahets
链接：https://leetcode.cn/problems/cong-shang-dao-xia-da-yin-er-cha-shu-lcof/solutions/134956/mian-shi-ti-32-i-cong-shang-dao-xia-da-yin-er-ch-4/
来源：力扣（LeetCode）
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