# Straightforward Iterative solution with O(1) space

• Use two pointers: one is the current node we are working on; the other is current node's parent. When we are at current node, we should have already completed assigning parent's next pointers, so we can move parent to its next node very easily.

Code in Java:

public class Solution {
public void connect(TreeLinkNode root) {
if(root==null) return;
TreeLinkNode parent = root; // current node's parent
TreeLinkNode current = root.left; // current node we are working on
TreeLinkNode next = root.right; // current node's next node
TreeLinkNode firstNode = current; // first node of every level
int leftOrRight = 1; // 1: current node is a left node; 2: current node is a right node.
while(current != null) {
if(leftOrRight==1) {
next = parent.right;
current.next = next;
current = next;
leftOrRight = 2; // current is now a right node
}
else {
if(parent.next != null) { // parent's next pointer is not null, thus moves to parent's next node
parent = parent.next;
next = parent.left;
current.next = next;
current = next;
}
else { // parent's next pointer is null, thus moves to next level
current.next = null;
parent = firstNode;
current = parent.left; // current moves to next level
firstNode = current; // first node of next level
}
leftOrRight = 1; // current is now a left node
}
}
}
}

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.