LeetCode 116. 填充每个节点的下一个右侧节点指针 JAVA

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。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

题解：利用层次遍历，将每一层的每一个结点都指向他的右边结点，
1、因为是完美二叉树，所以可以使用height来记录当前树的深度，然后通过计算2的height次方得到该层具有的节点数
2、对同一层的节点的next指针都指向它的右边结点，一层中的最后一个结点next指针置null

/*
// Definition for a Node.
class Node {
    public int val;
    public Node left;
    public Node right;
    public Node next;

    public Node() {}
    
    public Node(int _val) {
        val = _val;
    }

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

class Solution {
    public Node connect(Node root) {
        if(root==null) return null;
        Queue<Node> node=new LinkedList<>();
        int height=0;//高度
        int nodenum=1;//在某个高度下节点数
        node.offer(root);
        while(node.size()!=0)
        {
            nodenum--;
            Node tmp=node.poll();
            if(tmp.left!=null)
            {
                node.add(tmp.left);
                node.add(tmp.right);
            }
            if(nodenum==0)
            {
                height++;
                nodenum=(int)Math.pow(2,height);
            }
            else
            {
                if(node.size()!=0)
                {
                    Node next=node.peek();
                    tmp.next=next;
                }
            }
        }
        return root;
    }
}