Populating Next Right Pointers in Each Node

Problem

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

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */

Optimal DFS Solution without Extra Space

public class Solution {
    public void connect(TreeLinkNode root) {
        while(root != null) {
            TreeLinkNode node = root;
            while(node != null && node.left != null) {
                node.left.next = node.right;
                node.right.next = node.next == null ? null : node.next.left;
                node = node.next;
            }
            root = root.left;
        }
    }
}

More Intuitive BFS Solution using Queue: O(n) space

public class Solution {
    public void connect(TreeLinkNode root) {
        if (root == null) return;
        Queue<TreeLinkNode> q = new LinkedList<>();
        q.offer(root);
        while (!q.isEmpty()) {
            int size = q.size();
            List<TreeLinkNode> list = new ArrayList<>();
            for (int i = 0; i < size; i++) {
              TreeLinkNode node = q.poll();
              list.add(node);
              if (node.left != null) q.offer(node.left);
              if (node.right != null) q.offer(node.right);
            }
          for (int i = 0; i < size; i++) list.get(i).next = i == size - 1 ? null : list.get(i + 1);
        }
    }
}

Analysis

In the first solution we use DFS and two nested loops to set the next
For each node, we can set its left.next with its right
We set its right.next by checking whether its next == null, if so we set it to null, if not we set it to next.left
Since we are doing this from top to bottom and left to right, we cover all the nodes

The second solution is more intuitive by using BFS
We use a queue and its current size to get all nodes in same level
Then we connect nodes in the same level and continue to next level