算法题--连接二叉树同层节点II
2020-05-02 本文已影响0人
岁月如歌2020
image.png
0. 链接
1. 题目
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:
示意图
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
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.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-null-7=>null
7-null-null=>null
==================================================
1-2-3=>null
2-4-5=>3
4-null-null=>5
5-null-null=>7
3-null-7=>null
7-null-null=>null
4. 结果
image.png
5. 思路2: 三个指针+分层迭代
- 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':
head = pre = None
cur = root
while cur is not None:
while cur is not None:
if cur.left is not None:
if pre is not None:
pre.next = cur.left
else:
head = cur.left
pre = cur.left
if cur.right is not None:
if pre is not None:
pre.next = cur.right
else:
head = cur.right
pre = cur.right
cur = cur.next
cur = head
pre = head = None
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.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-null-7=>null
7-null-null=>null
==================================================
1-2-3=>null
2-4-5=>3
4-null-null=>5
5-null-null=>7
3-null-7=>null
7-null-null=>null
7. 结果
image.png
