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.
Example:
Input: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}
Output: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":{"$id":"6","left":null,"next":null,"right":null,"val":7},"right":null,"val":6},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"7","left":{"$ref":"5"},"next":null,"right":{"$ref":"6"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"7"},"val":1}
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.
Note:
- You may only use constant extra space.
- Recursive approach is fine, implicit stack space does not count as extra space for this problem.
Solutions
1. DFS-先序遍历-递归
采用递归的方法，注意这里的递归条件。
# Time Complexity: O(n)
# Space Complexity: O(logn)
"""
# Definition for a Node.
class Node:
def __init__(self, val, left, right, next):
self.val = val
self.left = left
self.right = right
self.next = next
"""
class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root:
return None
if not root.left:
return root
root.left.next = root.right
if root.next:
root.right.next = root.next.left
root.left = self.connect(root.left)
root.right = self.connect(root.right)
return root
# Runtime: 60 ms, faster than 98.51% of Python3 online submissions for Populating Next Right Pointers in Each Node.
# Memory Usage: 14.8 MB, less than 100.00% of Python3 online submissions for Populating Next Right Pointers in Each Node.
2. BFS-迭代
# Time Complexity: O(n)
# Space Complexity: O(logn)
"""
# Definition for a Node.
class Node:
def __init__(self, val, left, right, next):
self.val = val
self.left = left
self.right = right
self.next = next
"""
class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root:
return None
queue = collections.deque([root])
while queue:
size = len(queue)
pre = None
for i in range(size):
node = queue.popleft()
if not pre:
pre = node
else:
pre.next = node
pre = node
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return root
# Runtime: 64 ms, faster than 93.22% of Python3 online submissions for Populating Next Right Pointers in Each Node.
# Memory Usage: 14.8 MB, less than 100.00% of Python3 online submissions for Populating Next Right Pointers in Each Node.