Populating Next Right Pointers in Each Node II

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

------------------------- thinking -----------------------------------
using two level traversal methods
------------------------- codes -----------------------------------------
class Solution {
public:
    void connect(TreeLinkNode *root) {
        if (root == NULL) {
            return;
        }
        TreeLinkNode *cur = root;
        cur->next = NULL;
        TreeLinkNode *nxt_top = NULL;
        TreeLinkNode *nxt_cur = NULL;
        while (cur) {
            if (cur->left) {
                if (nxt_top == NULL) {
                    nxt_top = cur->left;
                    nxt_cur = cur->left;
                } else {
                    nxt_cur->next = cur->left;
                    nxt_cur = cur->left;
                }
            }
            if (cur->right) {
                if (nxt_top == NULL) {
                    nxt_top = cur->right;
                    nxt_cur = cur->right;
                } else {
                    nxt_cur->next = cur->right;
                    nxt_cur = cur->right;
                }
            }
            cur = cur->next;
            if (cur == NULL) {
                cur = nxt_top;
                nxt_top = nxt_cur = NULL;
            }
        }
    }
};
----------------------------- read codes ------------------------------------------------------
public class Solution { public void connect(TreeLinkNode root) { while(root != null){ TreeLinkNode tempChild = new TreeLinkNode(0); TreeLinkNode currentChild = tempChild; while(root!=null){ if(root.left != null) { currentChild.next = root.left; currentChild = currentChild.next;} if(root.right != null) { currentChild.next = root.right; currentChild = currentChild.next;} root = root.next; } root = tempChild.next; } } }