116. Populating Next Right Pointers in Each Node

Given a binary tree

    struct TreeLinkNode {
      TreeLinkNode *left;
      TreeLinkNode *right;
      TreeLinkNode *next;
    }

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

• You may only use constant extra space.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,
Given the following perfect binary tree,

         1
       /  \
      2    3
     / \  / \
    4  5  6  7

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

Solution:

We use level order traversal and maintain the start node in each level.

Since this is a perfect binary tree, every node except the leaves has both left and right children.

Therefore, for each node we can connect its left child with right child. (curr.left.next = curr.right)

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

For a right child, if curr is not the last node in the level, we can connect the right child with curr node's neighbor's left child. (curr.right.next = curr.next.left)

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

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 levelStart = root;
        
        while (levelStart != null) {
            TreeLinkNode curr = levelStart;
            
            while (curr != null) {
                if (curr.left != null) {
                    curr.left.next = curr.right;
                }
                if (curr.right != null && curr.next != null) {
                    curr.right.next = curr.next.left;
                }
                curr = curr.next;
            }
            
            levelStart = levelStart.left;
        }
    }
}