116. 填充每个节点的下一个右侧节点指针

填充每个节点的下一个右侧节点指针

示例：

输入：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}

输出：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":{"$id":"6","left":null,"next":null,"right":null,"val":7},"right":null,"val":6},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"7","left":{"$ref":"5"},"next":null,"right":{"$ref":"6"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"7"},"val":1}

解释：给定二叉树如图 A 所示，你的函数应该填充它的每个 next 指针，以指向其下一个右侧节点，如图 B 所示。

思路+代码+注释：

static class Node {
        public int val;
        public Node left;
        public Node right;
        public Node next;

        public Node() {}

        public Node(int _val,Node _left,Node _right,Node _next) {
            val = _val;
            left = _left;
            right = _right;
            next = _next;
        }
    }

    public static void main(String[] args) {
        Node n1=new Node();
        n1.val=1;
        Node n2=new Node();
        n2.val=2;
        Node n3=new Node();
        n3.val=3;
        Node n4=new Node();
        n4.val=4;
        Node n5=new Node();
        n5.val=5;
        Node n6=new Node();
        n6.val=6;
        Node n7=new Node();
        n7.val=7;
        n1.left=n2;
        n1.right=n3;
        n2.left=n4;
        n2.right=n5;
        n3.left=n6;
        n3.right=n7;
        connect(n1);
    }

    public static Node connect(Node root) {
            /*
            思路：
            1. 创建队列queue，初始装有根节点，弹出队列的节点加到list中，并且将弹出节点的左右子节点添加到queue中，直到queue为空
            2. 计算list中的节点个数，根据2aN-1=节点数，求出aN，根据aN求出最大层数
            3. 遍历每一层节点，每一层节点数为2^当前层数，每一层除了最后一个节点外其他节点的next指向下一个节点
             */
            if (root==null)
            {
                return null;
            }
            Queue<Node> queue=new LinkedList<>();
            ((LinkedList<Node>) queue).add(root);
            List<Node> nodeList=new ArrayList<>();
            while (queue.size()>0)
            {
             Node node=((LinkedList<Node>) queue).pop();
             nodeList.add(node);
             if (node.left!=null)
             {
                 ((LinkedList<Node>) queue).add(node.left);
             }
             if (node.right!=null)
             {
                    ((LinkedList<Node>) queue).add(node.right);
             }
            }
            int nodeNum=nodeList.size();
            int aN=(nodeNum+1)/2;
            int floorNum=0;
            while (aN!=1)
            {
                aN/=2;
                floorNum++;
            }
            int pos=0;
        for (int i = 0; i <= floorNum; i++) {
            double curFloorNum=Math.pow(2,i);
            for (int j = 0; j < curFloorNum; j++) {
                if (j==curFloorNum-1)
                {
                    nodeList.get(pos).next=null;
                }else {
                    nodeList.get(pos).next=nodeList.get(pos+1);
                }
                pos++;
            }
        }
        return root;
    }