[LeetCode]116. Populating Next Right Pointers in Each Node

这个题，我称作：找兄弟，next就是兄弟

要弄清楚家族里辈分关系

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

理解题意以后，想清楚节点之间的关系，左节点和右节点是两种情况，左节点的next一定是他亲兄弟，右节点的next是他堂兄弟，由于二叉树只能向下遍历，所以要在父辈中传入左右孩子需要的next

public void connect(TreeLinkNode root) {
        help(root,null);
    }
    public void help(TreeLinkNode root,TreeLinkNode right)
    {
        if (root==null)
            return;
        root.next = right;
        help(root.left,root.right);
        if (right!=null)
        {
            help(root.right,right.left);
        }
        else help(root.right,null);
    }

 第二部分：

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

• You may only use constant extra space.

For example,
Given the following binary tree,

         1
       /  
      2    3
     /     
    4   5    7

After calling your function, the tree should look like:

         1 -> NULL
       /  
      2 -> 3 -> NULL
     /     
    4-> 5 -> 7 -> NULL
这个题相对于上一问，改变的就是树可能不是perfect Tree,所以在找next的时候，左右节点都有可能是很远的“亲戚”，比如你叔叔没有孩子，那么你的next就是你二爷爷的孙子，甚至你二太爷爷的曾孙，等等...
所以在找next的时候要递归或者迭代实现，找到以后直接传递下去就行
我的悟性太差了，一开始发现这问题以后只是多判断了一下二爷爷的孙子，但是没有想到可能是更远的，后来错了才发现应该迭代。
这个问题也可以用迭代实现

public void connect(TreeLinkNode root) {
        help(root,null);
    }
    public void help(TreeLinkNode root,TreeLinkNode right)
    {
        if (root==null) return;
        root.next = right;
        TreeLinkNode next = finder(right);
        if (root.right!=null)
        {
            help(root.right,next);
            help(root.left,root.right);
        }
        else help(root.left,next);
    }
    public TreeLinkNode finder(TreeLinkNode node)
    {
        if (node!=null)
        {
            if (node.left!=null) return node.left;
            else if (node.right!=null) return node.right;
            else  return finder(node.next);
        }
        else return null;
    }