Populating Next Right Pointers in Each Node II <leetcode>

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

算法：该题和第一个题的变种，不过揭发是一样的，首先也是只要考虑当前节点，然后递归就行了，要点有二：1、当该节点左子树或右子树的右边的第一个节点   2、递归时要先算右子树，因为左子树的处理递归时会用到右边已经形成的结果，代码如下：

/**
 * 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) {
        if(NULL==root)  return;
        if(NULL!=root->left)
        {
             if(NULL!=root->right)
             {
                 root->left->next=root->right;
             }
             else  root->left->next=find(root->next);
        }
        if(NULL!=root->right)
        {
            root->right->next=find(root->next);
        }
        
        
        connect(root->right);
        connect(root->left);
    }
    
    TreeLinkNode* find(TreeLinkNode *temp)
    {
        if(NULL==temp)  return NULL;
        while(NULL!=temp)
        {
            if(NULL!=temp->left)  return temp->left;
            else if(NULL!=temp->right)  return temp->right;
            else temp=temp->next;
        }
        return NULL;
    }
};