116. 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

这个题想法还是比较单纯简单的，因为是完全二叉树，所以先将二叉树整体最右边一列的next赋值为NULL，然后考虑根节点两棵子树之间的关系，递归。
代码：

void right_null(TreeLinkNode* root)
{
    root->next=NULL;
    if (root->right==NULL)
        return;
    right_null(root->right);
}

void mid_next(TreeLinkNode* lefts,TreeLinkNode* rights)
{
    lefts->next=rights;
    TreeLinkNode* le=lefts;
    TreeLinkNode* ri=rights;
    while(le->right!=NULL)
    {
        le->right->next=ri->left;
        le=le->right;
        ri=ri->left;
    }
    if (lefts->left==NULL)
        return;
    mid_next(lefts->left,lefts->right);
    mid_next(rights->left,rights->right);
}

void connect(TreeLinkNode *root)
{
    if(root==NULL)
        return;
    right_null(root);
    if (root->left==NULL)
        return;
    mid_next(root->left,root->right);

}