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

Analysis

• 和之前一题不同之处：可以使不perfect的tree。前一题的code如下：

public class Solution {
    public void connect(TreeLinkNode root) {
        if (root==null) {return;}
        if (root.left!=null){
            root.left.next = root.right;
            if (root.next!=null){
                root.right.next = root.next.left;
            }
        }
        connect(root.left);
        connect(root.right);
    }
}

Example:
         1
       /  \
      2    3
     / \    \
    4   5    7
`

Code

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void connect(TreeLinkNode root) {
        TreeLinkNode head = null; // head of the next level
        TreeLinkNode prev = null; //the leading node on the next level
        TreLinkNode cur = root; //current node of current level

        while (cur != null){
            //iterate on the current level
            while (cur!=null){
                //left child
                if (cur.left != null){
                    if (prev != null){
                        prev.next = cur.left;
                    }
                    else {
                        head = cur.left;
                    }
                    prev = cur.left;
                }

                //right child
                if (cur.right !=null){
                    if (prev != null){
                        prev.next = cur.right;
                    }
                    else {
                        prev = cur.right;
                    }
                    prev = cur.right;
                }
                //move to next node
                cur = cur.next;
            }

            //move to the next level
            cur = head;
            head = null;
            prev = null;
        }

    }
}

Reference

Leetcode