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

 刚开始想到用BFS的做法，逐层遍历，在每一层中，将当前的节点的next指针指向下一个节点，这种做法只对完全二叉树有效
 对于如下的无效

             1
           /  \
          2    3       
         /    / \
        4     6  7

 得到的答案是
             1 -> NULL
           /  \
          2 -> 3 -> NULL
         /    / \
        4   ->6->7 -> NULL
  而正确的答案是:      
              1 -> NULL
           /     \
          2   ->  3 -> NULL
         /       / \
        4->NULL 6->7 -> NULL

 /**
 * 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(root == NULL)    return;
        TreeLinkNode *pre = root, *cur = NULL;

        while(pre->left){
            cur = pre;
            //将cur的左右儿子节点以及位于同一层的节点next指针赋值
            while(cur){
                cur->left->next = cur->right;
                if(cur->next)   cur->right->next = cur->next->left;
                cur = cur->next;
            }
            pre = pre->left;
        }
    }
};

递归的版本

 /**
 * 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(root == 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);
    }
};