手撕算法-二叉树的层序遍历

描述

分析

层序遍历，需要用到队列。

代码

代码1：独立bfs函数

/**
 * 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 {
    List<List<Integer>> res = new ArrayList<>();

    public List<List<Integer>> levelOrder(TreeNode root) {
        if (root == null)
            return res;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        bfs(queue);
        return res;
    }

    public void bfs(Queue<TreeNode> queue) {
        // 记录这一层的节点数
        int len = queue.size();
        if (len == 0)
            return;

        List<Integer> list = new ArrayList<>();
        // 遍历这一层的所有节点
        for (int i = 0; i < len; i++) {
            TreeNode node = queue.poll();
            list.add(node.val);
            if (node.left != null)
                queue.offer(node.left);
            if (node.right != null)
                queue.offer(node.right);
        }

        res.add(list);
        bfs(queue);
    }
}

代码2：无独立函数，更简洁

/**
 * 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>> res = new ArrayList<>();
        Queue<TreeNode> queue = new LinkedList<>();
        if (root != null)
            queue.offer(root);
        while (!queue.isEmpty()) {
            List<Integer> list = new ArrayList<>();
            for (int i = queue.size(); i > 0; i--) {
                TreeNode node = queue.poll();
                list.add(node.val);
                if (node.left != null)
                    queue.offer(node.left);
                if (node.right != null)
                    queue.offer(node.right);
            }
            res.add(list);
        }
        return res;
    }
}