题意

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。

提示：

你只能使用常量级额外空间。使用递归解题也符合要求，本题中递归程序占用的栈空间不算做额外的空间复杂度。思路

题目要求使用O（1）的额外空间，所以考虑类似BFS的算法。

因为树是完美的，那么当前这一层和上一层的关系是紧密的，体现在上一层节点cur存在next不为null那么当前层cur.left也存在next并且cur.right也存在next，可以根据示例图理解。每一层从上一层的最左边节点的左孩子开始遍历。

代码

/*// 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) { Node pre=root; while(pre!=null){ Node cur=pre; while(cur!=null){ if(cur.left!=null) cur.left.next=cur.right; if(cur.right!=null&&cur.next!=null){ cur.right.next=cur.next.left; } cur=cur.next; } pre=pre.left; } return root; }}