image.png
0. 链接
1. 题目
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:
示意图
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
2. 思路1: 队列+分层迭代
- 时间复杂度: ```O(N)``
- 空间复杂度:
O(N)
3. 代码
# coding:utf8
# 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 connect(self, root: 'Node') -> 'Node':
if root is None:
return None
queue = list()
queue.append(root)
while len(queue) > 0:
size = len(queue)
pre = None
for i in range(size):
each_node = queue.pop(0)
if pre is not None:
pre.next = each_node
pre = each_node
if each_node.left is not None:
queue.append(each_node.left)
if each_node.right is not None:
queue.append(each_node.right)
return root
def print_tree(node):
if node is None:
return
left = str(node.left.val) if node.left is not None else 'null'
right = str(node.right.val) if node.right is not None else 'null'
next = str(node.next.val) if node.next is not None else 'null'
print('{}-{}-{}=>{}'.format(node.val, left, right, next))
if node.left is not None:
print_tree(node.left)
if node.right is not None:
print_tree(node.right)
solution = Solution1()
root1 = node = Node(1)
node.left = Node(2)
node.right = Node(3)
node.left.left = Node(4)
node.left.right = Node(5)
node.right.left = Node(6)
node.right.right = Node(7)
print_tree(root1)
print('=' * 50)
solution.connect(root1)
print_tree(root1)
输出结果
1-2-3=>null
2-4-5=>null
4-null-null=>null
5-null-null=>null
3-6-7=>null
6-null-null=>null
7-null-null=>null
==================================================
1-2-3=>null
2-4-5=>3
4-null-null=>5
5-null-null=>6
3-6-7=>null
6-null-null=>7
7-null-null=>null
4. 结果
image.png
5. 思路2: 递归
- 时间复杂度:
O(N) - 空间复杂度:
O(1)
6. 代码
# coding:utf8
# 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 connect(self, root: 'Node') -> 'Node':
if root is None:
return
if root.left is not None:
root.left.next = root.right
if root.next is not None:
root.right.next = root.next.left
self.connect(root.left)
self.connect(root.right)
return root
def print_tree(node):
if node is None:
return
left = str(node.left.val) if node.left is not None else 'null'
right = str(node.right.val) if node.right is not None else 'null'
next = str(node.next.val) if node.next is not None else 'null'
print('{}-{}-{}=>{}'.format(node.val, left, right, next))
if node.left is not None:
print_tree(node.left)
if node.right is not None:
print_tree(node.right)
solution = Solution()
root1 = node = Node(1)
node.left = Node(2)
node.right = Node(3)
node.left.left = Node(4)
node.left.right = Node(5)
node.right.left = Node(6)
node.right.right = Node(7)
print_tree(root1)
print('=' * 50)
solution.connect(root1)
print_tree(root1)
7. 结果
image.png
