Tree - 116. Populating Next Right Pointers in Each Node

116. Populating Next Right Pointers in Each Node

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.

**思路：**
> 给一棵满二叉树，让把next指针指向同一层，很简单，可以先序遍历做，也可以只用两个指针，核心代码就是对于一个节点，左节点指向右节点，右节点指向自己的下一个的左节点。因为题目要求`only use constant extra space`,所以不能用递归，递归其实也是用了额外的空间。

**代码：**

java：
/*
// 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) {
 recursion(root);
 return root;
 }

private void recursion(Node root) {
    if (root == null) return;
    
    if (root.left != null) root.left.next = root.right;
    if (root.right != null && root.next != null) root.right.next = root.next.left;
    
    recursion(root.left);
    recursion(root.right);
}*/

public Node connect(Node root) {
    Node level_start = root;
    
    while (level_start != null) {
        Node cur = level_start;
        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;
        }
        level_start = level_start.left;
    }
    
    return root;
 }
}