Populating Next Right Pointers in Each Node II
Created: February 26, 2020 by [lek-tin]
Last updated: February 26, 2020
Given a binary tree
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,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.
Constraints:
- The number of nodes in the given tree is less than
6000. -100 <= node.val <= 100
Solution 1 (bfs / level by level)
Time: O(n)
Space: O(n)
"""
# Definition for a Node.
class Node:
def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
self.val = val
self.left = left
self.right = right
self.next = next
"""
from collections import deque
class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root:
return root
queue = deque([root])
while queue:
size = len(queue)
node = queue[0]
for i in range(size):
node = queue.popleft()
if i < size-1:
node.next = queue[0]
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return root
Solution 2 (iterative)
Time: O(n)
Space: O(1)
"""
# Definition for a Node.
class Node:
def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
self.val = val
self.left = left
self.right = right
self.next = next
"""
from collections import deque
class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root:
return root
head_nextLevel = None
prev_nextLevel = None
curr = root
while curr:
while curr:
if curr.left:
if head_nextLevel:
prev_nextLevel.next = curr.left
prev_nextLevel = prev_nextLevel.next
else:
head_nextLevel = curr.left
prev_nextLevel = curr.left
if curr.right:
if head_nextLevel:
prev_nextLevel.next = curr.right
prev_nextLevel = prev_nextLevel.next
else:
head_nextLevel = curr.right
prev_nextLevel = curr.right
curr = curr.next
curr = head_nextLevel
head_nextLevel = None
prev_nextLevel = None
return root