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

一刷
题解：题目特点，假设input是一个perfect binary tree. 一层一层降下去，在上一层，连接它的左右。在当前层，iterate这个层的每个node, 连接它们的左右。

/**
 * 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) {
        TreeLinkNode level_start = root;
        while(level_start != null){
            TreeLinkNode 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;
        }
    }
}

二刷：
思路同上。

/**
 * 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) {
        //preorder
        if(root == null || root.left==null) return;
        TreeLinkNode level = root;
        TreeLinkNode cur = root;
        while(level!=null && level.left!=null){
            cur = level;
            while(cur!=null && cur.left!=null){
                cur.left.next = cur.right;
                if(cur.next!=null && cur.right!=null){
                    cur.right.next = cur.next.left;
                }
                cur = cur.next;
            }
            level = level.left;
        }
    }
}

三刷
逐级下沉

/**
 * 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;
        TreeLinkNode next = root.left;
        while(root!=null && next!=null){
            root.left.next = root.right;
            if(root.next!=null) root.right.next = root.next.left;
            root = root.next;
        }
        connect(next);
    }
}