Leetcode116. 填充每个节点的下一个右侧节点指针

题目传送： ​​https://leetcode.cn/problems/populating-next-right-pointers-in-each-node/​​

运行效率

代码如下：

class Solution {
   public Node connect(Node root) {
        //处理边界情况
        if(root==null||(root.left==null&&root.right==null)){
            return root;
        }
        List<Node> list = getList(root);
        for (int i=0;i<list.size()-1;i++){
            Node curNode = list.get(i);
            Node nextNode = list.get(i + 1);
            if (curNode!=null){
                curNode.next=nextNode;
            }
        }
        return list.get(0);
     
    }


    public List<Node> getList(Node root) {
        List<Node> result = new ArrayList<>();
        //层序遍历一般通过队列来实现
        ArrayDeque<Node> deque = new ArrayDeque<>();
        deque.add(root);
        while (!deque.isEmpty()) {
            //用于存储下一层的所有节点
            ArrayDeque<Node> arrayDeque = new ArrayDeque<>();
            while (!deque.isEmpty()) {
                Node poll = deque.poll();
                result.add(poll);
                if (poll.left != null) {
                    arrayDeque.add(poll.left);
                }
                if (poll.right != null) {
                    arrayDeque.add(poll.right);
                }
            }
            result.add(null);
            deque.addAll(arrayDeque);
        }
        return result;
    }

}