leetcode：116. 填充每个节点的下一个右侧节点指针（java树层次遍历bfs）

package LeetCode;

import java.util.LinkedList;
import java.util.Queue;
/*
给定一个完美二叉树，其所有叶子节点都在同一层，每个父节点都有两个子节点。二叉树定义如下：
struct Node {
  int val;
  Node *left;
  Node *right;
  Node *next;
}
填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。
初始状态下，所有 next 指针都被设置为 NULL。
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。
 */
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 class Connect {
    /*
    思路：用层次遍历然后先把第一个取出然后遍历剩余的节点时用next连起来即可
     */
    public Node connect(Node root) {
        if(root==null) return null;
        Queue<Node> queue = new LinkedList<>();
        queue.add(root);
        while (!queue.isEmpty()) {
            int qcount = queue.size();
            qcount--;
            Node first = queue.poll();
            if (first.left != null) {
                queue.add(first.left);
            }
            if (first.right != null) {
                queue.add(first.right);
            }
            while (qcount > 0) {
                Node now = queue.poll();
                if (now.left != null) {
                    queue.add(now.left);
                }
                if (now.right != null) {
                    queue.add(now.right);
                }
                first.next = now;
                first = first.next;
                qcount--;
            }
        }
        return root;
    }

    public static void main(String[] args) {
        Node node4 = new Node(4, null, null, null);
        Node node5 = new Node(5, null, null, null);
        Node node6 = new Node(6, null, null, null);
        Node node7 = new Node(7, null, null, null);
        Node node2 = new Node(2, node4, node5, null);
        Node node3 = new Node(3, node6, node7, null);
        Node node1 = new Node(1, node2, node3, null);
        Connect connect = new Connect();
        connect.connect(node1);

    }
}