Populating Next Right Pointers in Each Node II
Problem Description
Given a binary tree
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.
Example 1:
Input: root = [1,2,3,4,5,null,7]
Output: [1,#,2,3,#,4,5,7,#]
Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.
Example 2:
Input: root = []
Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 6000]. -100 <= Node.val <= 100
Follow-up:
- You may only use constant extra space.
- The recursive approach is fine. You may assume implicit stack space does not count as extra space for this problem.
Solve it on LeetCode
Approaches
1. Level Order Traversal using Queue
Intuition:
The basic idea is to use a queue to perform a level order traversal of the tree. For each node at a particular level, connect it to its next right node using the queue. This method ensures that we connect all nodes at the same level before moving to the next level.
- Start by enqueuing the root node.
- Iterate while the queue is not empty. For every iteration, it represents a new level.
- For each node at the current level, link node's
nextto the next node in the queue. - Enqueue the left and right children of the node if they exist.
- Repeat the above steps until all levels are processed.
Code:
- Time Complexity: O(N), where N is the total number of nodes. Each node is processed once.
- Space Complexity: O(N), the space used by the queue in the worst case.
2. Using Previously Established Next Pointers
Intuition:
To optimize space, we can avoid using a queue and instead leverage the next pointers established during the traversal. Instead of processing nodes level by level, the idea is to build the next pointers for the next level while traversing the current level.
- Use a current pointer to traverse nodes at the current level.
- Use a dummy node to represent the start of the next level.
- Track the last visited node in the next level using a tail pointer starting from the dummy node.
- Move to the next level by setting the current pointer to
dummy.nextonce the current level is completely processed.
Code:
- Time Complexity: O(N), where N is the total number of nodes. Each node is processed once.
- Space Complexity: O(1), as we're only using a few additional pointers, not a queue.