leetcode-116-填充每个节点的下一个右侧节点指针

原文链接：http://www.cnblogs.com/oldby/p/11185617.html

题目描述：

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。

方法一：递归：

class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root:
return
if root.left:
root.left.next = root.right
if root.next:
root.right.next = root.next.left
self.connect(root.left)
self.connect(root.right)
return root

方法二：迭代：

class Solution:
def connect(self, root: 'Node') -> 'Node':
pre = root
while pre:
cur = pre
while cur:
if cur.left: cur.left.next = cur.right
if cur.right and cur.next:
cur.right.next = cur.next.left
cur = cur.next
pre = pre.left
return root

 

 

转载于:https://www.cnblogs.com/oldby/p/11185617.html