116. Populating Next Right Pointers in Each Node

Problem:

Given a binary tree

    struct TreeLinkNode {
      TreeLinkNode *left;
      TreeLinkNode *right;
      TreeLinkNode *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.

Note:

• You may only use constant extra space.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,
Given the following perfect binary tree,

         1
       /  \
      2    3
     / \  / \
    4  5  6  7

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

4/24/2018 update:
这道题感觉iterative要好很多，没有什么递归的必要。
Iterative:

public class Solution {
    public void connect(TreeLinkNode root) {
        TreeLinkNode start = root;
        while (start != null) {
            TreeLinkNode cur = start;
            while (cur != null) {
                if (cur.left != null) {
                    cur.left.next = cur.right;
                    if (cur.next != null) {
                        cur.right.next = cur.next.left;
                    }
                }
                cur = cur.next;
            }
            start = start.left;
        }   
    }
}

Recursive:
递归实现来说要简单很多，但是感觉这是假递归。因为递归的结果并不影响当前node。

public class Solution {
    public void connect(TreeLinkNode root) {
        if (root == null) return;

        if (root.left != null) {
            root.left.next = root.right;
            if (root.next != null)
                root.right.next = root.next.left;
        }
        
        connect(root.left);
        connect(root.right);
    }
}

--------------------------------------------------------------------------------------------------------------------------
Soluiton:

 /**  
  * Definition for binary tree with next pointer.  
  * public class TreeLinkNode {  
  *   int val;  
  *   TreeLinkNode left, right, next;  
  *   TreeLinkNode(int x) { val = x; }  
  * }  
  */  
 public class Solution {  
   public void connect(TreeLinkNode root) {  
     if (root == null) return;  
     if (root.left != null) {  
       root.left.next = root.right;  
       if (root.next != null)  
         root.right.next = root.next.left;  
     }  
     connect(root.left);  
     connect(root.right);  
   }  
 }