力扣hot100 二叉树的层序遍历 BFS 队列

‍ 题目地址

• 时间复杂度：$O(n)$
• 空间复杂度：$O(n)$

队列写法

/**
 * 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<>();
//当前层的结点出队完的时候，队列中的结点只可能是下一层的，下下层的结点当前层够不着
		LinkedList<TreeNode> q = new LinkedList<>();//只存当前层的结点

		if (root == null)
			return ans;
		q.add(root);
		while (!q.isEmpty())
		{
			int n = q.size();//n记录当前层的结点个数
			ArrayList<Integer> list = new ArrayList<>(n);
			while (n-- > 0)
			{
				TreeNode node = q.poll();
				list.add(node.val);
				if (node.left != null)
					q.add(node.left);
				if (node.right != null)
					q.add(node.right);
			}
			ans.add(list);
		}
		return ans;
	}
}

顺序表写法

class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        if (root == null) return List.of();
        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()); // 容量已知
            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;
    }
}

‍ 参考题解