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
     / \    \
    4   5    7

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \    \
    4-> 5 -> 7 -> NULL

 

Analyse: BFS. Generally, we use queue to do BFS. However, in this problem, we can use the "next pointer" to point to the next node in the same level. When we consider the current level, actually we are dealing with the next level. The structure of the next level likes a linked list. 

Runtime: 40ms.

/**
 * 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) return;
        
        TreeLinkNode pre(-1); //previous node in the level
        for(TreeLinkNode* current = root, *move = ⪯ current; current = current->next){
            if(current->left){
                move->next = current->left;
                move = move->next;
            }
            if(current->right){
                move->next = current->right;
                move = move->next;
            }
            //current = current->next; //children of the current node have been dealt, move to the next node in the level
        }
        root = pre.next; //assign the first node in the next level to root and begin the new recursion
        connect(root);
    }
};