一、题目
1、LeetCode116:
给定一个完美二叉树,其所有叶子节点都在同一层,每个父节点都有两个子节点。二叉树定义如下:
struct Node {
int val;
Node *left;
Node *right;
Node *next;
}
填充它的每个 next 指针,让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点,则将 next 指针设置为 NULL。
初始状态下,所有 next 指针都被设置为 NULL。
2、LeetCode117:
给定一个二叉树
struct Node {
int val;
Node *left;
Node *right;
Node *next;
}
填充它的每个 next 指针,让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点,则将 next 指针设置为 NULL。初始状态下,所有 next 指针都被设置为 NULL。
进阶:
- 你只能使用常量级额外空间。
- 使用递归解题也符合要求,本题中递归程序占用的栈空间不算做额外的空间复杂度。
二、示例
1、LeetCode116:
示例:
输入:{"$id":"1","left":{"$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":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}
输出:{"$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}
解释:给定二叉树如图 A 所示,你的函数应该填充它的每个 next 指针,以指向其下一个右侧节点,如图 B 所示。
提示:
- 你只能使用常量级额外空间。
- 使用递归解题也符合要求,本题中递归程序占用的栈空间不算做额外的空间复杂度。
2、LeetCode117:
示例:
输入:root = [1,2,3,4,5,null,7]
输出:[1,#,2,3,#,4,5,7,#]
解释:给定二叉树如图 A 所示,你的函数应该填充它的每个 next 指针,以指向其下一个右侧节点,如图 B 所示。
提示:
- 树中的节点数小于 6000
- -100 <= node.val <= 100
三、思路
1、LeetCode116:
注:方法2,3均可适用LeetCode117
(1)题意是将每一层的结点连接起来,每个结点的next指针均指向右边的结点。
- 如果右边结点和自己是同一个父节点,root.left.next = root.right,即B中2-->3
- 如果右边结点和自己不是同一个父节点,即堂兄弟的关系,root.right.next = root.next.left,即图B中5-->6
(2)层次遍历+队列
(3)每一行都可以看成一个链表。
比如第一行就是只有一个节点的链表,第二行是只有两个节点的链表(假如根节点的左右两个子节点都不为空)……
2、LeetCode117:
(1)同上述2
(2)同上述3
四、代码
(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
def connect(self, root: 'Node'):
if root is None:
return
if root.left:
root.left.next = root.right
if root.right and root.next:
root.right.next = root.next.left
self.connect(root.left)
self.connect(root.right)
return root
if __name__ == '__main__':
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.right = Node(7)
s = Solution()
ans = s.connect(root)
print(ans)
(2)
# 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'):
"""
:type root: Node
:rtype: Node
"""
if root is Node:
return
queue = [root]
while queue:
n = len(queue)
for i in range(n):
t = queue.pop(0)
if t.left:
queue.append(t.left)
if t.right:
queue.append(t.right)
if i < n - 1:
t.next = queue[0]
return root
if __name__ == '__main__':
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.right = Node(7)
s = Solution()
ans = s.connect(root)
print(ans)
(3)
# 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'):
if root is Node:
return
cur = root
while cur:
dummy = Node(0)
pre = dummy
while cur:
if cur.left:
pre.next = cur.left
pre = pre.next
if cur.right:
pre.next = cur.right
pre = pre.next
cur = cur.next
cur = dummy.next
return root
if __name__ == '__main__':
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.right = Node(7)
s = Solution()
ans = s.connect(root)
print(ans)