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.
Example 1:
1 1--> NULL
/ \ / \
2 3 2-->3--> NULL
/ \ \ / \ \
4 5 7 4-->5-->7--> NULL
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.
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"""
# 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
"""
class Solution:
def process_child(self, child, prev, leftmost):
if child:
if prev:
prev.next = child
else:
leftmost = child
prev = child
return prev, leftmost
def connect(self, root: 'Node') -> 'Node':
if not root:
return root
leftmost = root
while leftmost:
prev, cur = None, leftmost
leftmost = None
while cur:
prev, leftmost = self.process_child(cur.left, prev, leftmost)
prev, leftmost = self.process_child(cur.right, prev, leftmost)
cur = cur.next
return root
'''
BFS
'''
class Solution:
def connect(self, root: 'Node') -> 'Node':
from collections import deque
if not root:
return root
q = deque([root])
while q:
size = len(q)
for i in range(size):
node = q.popleft()
if i < size - 1:
node.next = q[0]
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
return root
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