leetcode @python 116. Populating Next Right Pointers in Each Node

题目链接

https://leetcode.com/problems/populating-next-right-pointers-in-each-node/

题目原文

题目大意

二叉树的每个节点都包含左子树和右子树，将同一层的子树用链表连接起来

解题思路

递归求解：
1.如果该节点有左子树：1）该节点有右子树 2）该节点没有右子树 递归该节点的左子树
2.如果该节点有右子树：1）该节点的next节点有左子树 2）该节点的next节点没有左子树

代码

class TreeLinkNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
        self.next = None


class Solution(object):
    def connect(self, root):
        """
        :type root: TreeLinkNode
        :rtype: nothing
        """
        if not root:
            return
        else:
            root.next = None
            self.help(root)

    def help(self, node):
        if node.left:
            if node.right:
                node.left.next = node.right
            else:
                node.left.next = None
            self.help(node.left)
        if node.right:
            if node.next and node.next.left:
                node.right.next = node.next.left
            else:
                node.right.next = None
            self.help(node.right)