116. 填充每个节点的下一个右侧节点指针

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。

BFS模板的应用。

总结的DFS、BFS模板参考：https://www.cnblogs.com/panweiwei/p/13065661.html

class Solution(object):
    # 层序遍历
    def connect(self, root):
        """
        :type root: Node
        :rtype: Node
        """
        if not root:
            return root
        queue = [root]
        # while 循环迭代的是树的层级。
        while queue:
            l = len(queue)
            # 内部的 for 循环迭代的是同一层的所有节点
            for i in range(l):
                node = queue.pop(0)
                # 由于可以访问同一层级的所有节点，因此能够建立 next 指针连接。
                if i < l - 1:
                    node.next = queue[0]
                """
                for 循环弹出一个节点时，同时把它的左孩子节点和右孩子节点依次加入队列。
                因此队列中每个层级的元素也是顺序存储的。可以通过已有的顺序建立 next 指针。
                """
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
        return root