Leetcode-Populating Next Right Pointer in Binary Tree 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

Solution:

/**
 * 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) {
        if (root==null)
            return;
            
        root.next = null;
        TreeLinkNode levelHead = root;
        
        while (true){
            boolean hasChild = false;
            TreeLinkNode cur = levelHead;
            TreeLinkNode nextLevelPre = null;
            TreeLinkNode nextLevelHead = null;
            while (cur!=null){
                if (cur.left!=null){
                    hasChild = true;
                    if (nextLevelPre!=null){
                        nextLevelPre.next = cur.left;
                        nextLevelPre = cur.left;
                    } else {
                        nextLevelPre = cur.left;
                        nextLevelHead = cur.left;
                    }
                } 
                
                if (cur.right!=null){
                    hasChild = true;
                    if (nextLevelPre!=null){
                        nextLevelPre.next = cur.right;
                        nextLevelPre = cur.right;
                    } else {
                        nextLevelPre = cur.right;
                        nextLevelHead = cur.right;
                    }
                }
                
                cur = cur.next;
            }
            if (hasChild)
                nextLevelPre.next = null;
            
            //If reach the last level, then stop.
            if (!hasChild)
                break;
                
            //Move to next level.
            levelHead = nextLevelHead;
        }
        
        return;
    }
}

At each level, after we construct the link list, we have a linked list to visit all nodes in this level. Then we can visit all child nodes in the next level along this linked list.This is a very smart way.

Note: For this question, we need to be careful about where the first child node in the next level is.