Given a binary tree

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":{"$id":"5","left":null,"next":null,"right":{"$id":"6","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":{"$id":"5","left":null,"next":null,"right":null,"val":7},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"6","left":null,"next":null,"right":{"$ref":"5"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"6"},"val":1}Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B.

Note:

1. You may only use constant extra space.2. Recursive approach is fine, implicit stack space does not count as extra space for this problem.

难度：medium

题目：给定二叉树，计算结点指向其相邻右结点的下一跳指针。如果没有相邻的右结点则next为NULL.

思路：递归

Runtime: 3 ms, faster than 18.40% of Java online submissions for Populating Next Right Pointers in Each Node II.

/*// 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) {            ptr = buildNodeNext(ptr);        }                return root;    }        private Node buildNodeNext(Node head) {        if (null == head) {            return null;        }        Node nextLevelLeftMost = buildNodeNext(head.next);        Node left = head.left;        Node right = head.right;        if (left != null && right != null) {            left.next = right;            right.next = nextLevelLeftMost;            return left;        }                if (left != null && right == null) {            left.next = nextLevelLeftMost;            return left;        }                 if (left == null && right != null) {            right.next = nextLevelLeftMost;            return right;        }                return nextLevelLeftMost;    }}