[LeetCode]题解，python：116 Populating Next Right Pointers in Each Node

题目来源

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

Given a binary tree

    struct TreeLinkNode {
      TreeLinkNode *left;
      TreeLinkNode *right;
      TreeLinkNode *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.

Note:

• You may only use constant extra space.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,

Given the following perfect binary tree,

         1
       /  \
      2    3
     / \  / \
    4  5  6  7

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

题意分析

Input:满二叉树

Output:增加了next信息的满二叉树

Conditions:只能使用常量空间

题目思路

注意到是满二叉树，并且上一层的next信息可以用于下一层。

AC代码（Python）

# Definition for binary tree with next pointer.
# 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 root and root.left:
            root.left.next = root.right
            if root.next:
                root.right.next = root.next.left
            else:
                root.right.next = None
            self.connect(root.left)
            self.connect(root.right)