116. Populating Next Right Pointers in Each Node
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/
You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:
struct Node {
int val;
Node *left;
Node *right;
Node *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.
Follow up:
- You may only use constant extra space.
- Recursive approach is fine, you may assume implicit stack space does not count as extra space for this problem.
Example 1:
Input: root = [1,2,3,4,5,6,7] Output: [1,#,2,3,#,4,5,6,7,#] Explanation: Given the above perfect 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.
Constraints:
- The number of nodes in the given tree is less than
4096. -1000 <= node.val <= 1000
----
Intuition
Level Order BFS - if remaining > 0 set next pointer to q.poll()
---
We can use info that tree is complete to our advantage
root.left.next = root.right
if (root.next == null)
root.right.next = null
else
root.right.next = root.next.left
root = root.next -- to advance to next node on same level
This continues till root ! = null -- marking end of level
On next level, we need to start at leftmost node.. so save it before beginning above while loop .. root != null
levelStart = root.left
while (root != null)
....
once level is finished start at the saved node
root = levelStart
Continue this till root != null and root.left != null - complete binary tree => this is valid all levels except last level
---
Time - O(n)
Space - O(1)
---
Time - O(n)
Space - O(n) or O(2 ^ h - 1) - nodes of last level
---
Iterative solution is interesting---
We can use info that tree is complete to our advantage
root.left.next = root.right
if (root.next == null)
root.right.next = null
else
root.right.next = root.next.left
root = root.next -- to advance to next node on same level
This continues till root ! = null -- marking end of level
On next level, we need to start at leftmost node.. so save it before beginning above while loop .. root != null
levelStart = root.left
while (root != null)
....
once level is finished start at the saved node
root = levelStart
Continue this till root != null and root.left != null - complete binary tree => this is valid all levels except last level
---
Time - O(n)
Space - O(1)
