102. Binary Tree Level Order Traversal（二叉树的层序遍历）

题目描述

Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],

方法思路

class Solution {
    //Runtime: 1 ms, faster than 80.44%
    //Memory Usage: 37.4 MB, less than 35.53%
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if(root == null) return res;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        List<Integer> l = new ArrayList<>();
        l.add(root.val);
        res.add(l);
        while(!queue.isEmpty()){
            int size = queue.size();
            while(size-- > 0){
                TreeNode node = queue.poll();
                if(node.left != null) queue.offer(node.left);
                if(node.right != null) queue.offer(node.right);
            }
            if(queue.size() > 0){
                List<Integer> list = new ArrayList<>();
                for(TreeNode node : queue)
                    list.add(node.val);
                res.add(list);
            }
        }
        return res;
    }
}