题目一：Populating Next Right Pointers in Each Node

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

struct TreeLinkNode 
{
	int val;
    TreeLinkNode *left, *right, *next;
    TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
};

/* 由于题目说了是完全二叉树，所以如果由节点next不为空的话，那么一定是root -> next -> left */
class Solution {
public:
    void connect(TreeLinkNode *root) {
    	if(!root || (root -> left == NULL && root -> right == NULL))return;
    	if(root -> left)root -> left -> next = root -> right;
    	if(root -> right && root -> next) root -> right -> next = root -> next -> left;
    	connect(root -> left);
    	connect(root -> right);
    }
};

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

class Solution {
public:
    void connect(TreeLinkNode *root) {
    	if(!root || (root -> left == NULL && root -> right == NULL))return;	
    	TreeLinkNode* p ,*q;
    	if(root -> left)
    	{
    		if(root -> right)root -> left -> next = root -> right;
    		else
    		{
    			q = NULL;
    			p = root -> next;
    			while(p != NULL)//沿着父亲的next指针一直寻找
    			{
    				if(p -> left)
    				{
    					q = p -> left;
    					break;
    				}
    				else if(p -> right)
    				{
    					q = p -> right;
    					break;
    				}
    				p = p -> next;
    			}
    			root -> left ->  next = q;
    		}
    	}
    	if(root -> right)
    	{
    		q = NULL;
    		p = root -> next;
    		while(p != NULL)
    		{
    			if(p -> left)
    			{
    				q = p -> left;
    				break;
    			}
    			else if(p -> right)
    			{
    				q = p -> right;
    				break;
    			}
    			p = p -> next;
    		}
    		root -> right -> next = q;
    	}
    	connect(root -> right);//注意要先构造右子树
    	connect(root -> left);
    }

};