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
題解：題目特點旁钧，假設(shè)input是一個perfect binary tree. 一層一層降下去涉波，在上一層，連接它的左右爪瓜。在當(dāng)前層谊路，iterate這個層的每個node, 連接它們的左右赊锚。

/**
 * 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：
思路同上成箫。

/**
 * 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
逐級下沉

/**
 * 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);
    }
}