You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

struct Node {
  int val;
  Node *left;
  Node *right;
  Node *next;
}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.
Initially, all next pointers are set to NULL.
Example:

Input: {"$id":"1","left":{"$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}

Output: {"$id":"1","left":{"$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}

Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B.

Note:
You may only use constant extra space.
Recursive approach is fine, implicit stack space does not count as extra space for this problem.

难度：medium

题目：给定一满二叉树，所有叶结点在同一层，每个结点都有左右子树。

思路：自上至下自左至右遍历

Runtime: 0 ms, faster than 100.00% of Java online submissions for Populating Next Right Pointers in Each Node.
Memory Usage: 36.1 MB, less than 100.00% of Java online submissions for Populating Next Right Pointers in Each Node.

/*
// Definition for a Node.
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;
    }
};
*/
class Solution {
    public Node connect(Node root) {
        Node ptr = root;
        while (ptr != null) {
            Node head = ptr, rightMost = null;
            ptr = ptr.left;
            while (head != null && head.left != null) {
                if (rightMost != null) {
                    rightMost.next = head.left;
                }
                head.left.next = head.right;
                rightMost = head.right;
                head = head.next;
            }
        }
        
        return root;
    }
}