[LeetCode] 117. 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

Thought process:
Still level order traversal. But this time since the tree is not necessarily perfect, I can't iterate on root.left. Instead, I can use dummy nodes whose next pointer points to the left-most node of each level.

Solution 1 (BFS):
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/**
 * 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) {
        while (root != null) {
            TreeLinkNode dummy = new TreeLinkNode(0);
            TreeLinkNode current = dummy;
            
            while (root != null) {
                if (root.left != null) {
                    current.next = root.left;
                    current = current.next;
                }
                if (root.right != null) {
                    current.next = root.right;
                    current = current.next;
                }
                root = root.next;
            }
            root = dummy.next;
        }
    }
}

Solution 2 (DFS):

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/**
 * 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;
        }
        
        TreeLinkNode node = root.next;
        while (node != null) {
            if (node.left != null) {
                node = node.left;
                break;
            }
            if (node.right != null) {
                node = node.right;
                break;
            }
            node = node.next;
        }
        
        if (root.right != null) {
            root.right.next = node;
        }
        if (root.left != null) {
            root.left.next = root.right == null ? node : root.right;
        }
        
        connect(root.right);
        connect(root.left);
    }
}

Time complexity: O(n).
Location: Philadelphia, PA, USA

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