****（leetcode) Populating Next Right Pointers in Each Node II

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

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

1、根据完全二叉树方法修改判断语句即可！ 开始一直超时，跟着程序跑了一遍才发现一直conn2没有在每次都设为NULL。 思考不细致啊！！

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  TreeLinkNode *left, *right, *next;
 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 * };
 */
class Solution {
public:
    void connect(TreeLinkNode *root) {  
        TreeLinkNode *firstNodeEveryRow, *parent, *conn1,*conn2;  
        firstNodeEveryRow=root; //first..每一行的第一个节点  
        while(NULL != firstNodeEveryRow){  
            parent = firstNodeEveryRow;  
            conn1 = conn2=NULL; //conn1->next=conn2， 用于连接  
            while(NULL != parent){ //用于遍历改行的节点  
                if(parent->left != NULL&&parent->right != NULL)     //左右兄弟先连接  
                    parent->left->next = parent->right;  
                if(parent->left != NULL || parent->right != NULL)  
                    conn2 = (parent->left != NULL)?parent->left:parent->right;  
                if(conn1!=NULL&&conn2!=NULL){   //两个父节点之间的子节点相连接  
                     conn1->next=conn2;  
                     conn1=NULL;  
                }  
               
                if(conn1==NULL&&(parent->left!=NULL||parent->right!=NULL))  
                    conn1=(parent->right!=NULL)?parent->right:parent->left;  
                parent=parent->next;
                conn2=NULL; //!conn2表示的当前parent节点里面的用于和前面节点相连接的点，so parent更换后conn2要置为NULL，不然会成环！！
            }  
            while(firstNodeEveryRow != NULL && firstNodeEveryRow->left==NULL && firstNodeEveryRow->right==NULL)  
                firstNodeEveryRow=firstNodeEveryRow->next;  
            if(NULL != firstNodeEveryRow)  
                firstNodeEveryRow=(firstNodeEveryRow->left != NULL)?firstNodeEveryRow->left:firstNodeEveryRow->right;  
        }  }
};

2、递归：

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  TreeLinkNode *left, *right, *next;
 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 * };
 */
class Solution {
public:
    void connect(TreeLinkNode *root) {
        TreeLinkNode *rt=NULL, *nextNode=NULL;
        if(NULL==root)
            return;
        rt = root->next;
        while(rt!=NULL){
            if(rt->left!=NULL){
                nextNode = rt->left;
                break;
            }
            if(rt->right!=NULL){
                nextNode = rt->right;
                break;
            }
            rt=rt->next;
        }
        if(root->left!=NULL){
            if(root->right!=NULL)
                root->left->next = root->right;
            else
                root->left->next = nextNode;
        }
        if(root->right!=NULL){
            root->right->next = nextNode;
        }
        connect(root->right); //注意，先递归右侧，不然如果先递归左侧，右侧的还没有连接起来，左侧递归时无法连接
        connect(root->left);  // 因为左侧的连接需要右侧的信息
    }
};