leetcode/116.填充每个节点的下一个右侧节点指针.py at master · junncao/leetcode · GitHub

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#
# @lc app=leetcode.cn id=116 lang=python
#
# [116] 填充每个节点的下一个右侧节点指针
#
# https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node/description/
#
# algorithms
# Medium (48.22%)
# Likes:    94
# Dislikes: 0
# Total Accepted:    15.2K
# Total Submissions: 31.4K
# Testcase Example:  '{"$id":"1","left":{"$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":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}'
#
# 给定一个完美二叉树，其所有叶子节点都在同一层，每个父节点都有两个子节点。二叉树定义如下：
#
# struct Node {
# ⁠ int val;
# ⁠ Node *left;
# ⁠ Node *right;
# ⁠ Node *next;
# }
#
# 填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。
#
# 初始状态下，所有 next 指针都被设置为 NULL。
#
#
#
# 示例：
#
#
#
#
# 输入：{"$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。
#
#
#

# @lc code=start
"""
# Definition for a Node.
class Node(object):
    def __init__(self, val, left, right, next):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""
class Solution(object):
    def connect(self, root):
        """
        :type root: Node
        :rtype: Node
        """
        if not root or not root.left:return root  #为None 或者 为叶子节点
        root.left.next=root.right
        if root.next:
            root.right.next=root.next.left

        self.connect(root.left)
        self.connect(root.right)
                #要先处理根节点保证同层次之间已经串起来了才可以进行下一层，先左边还是右边的顺序无要求
        return root
# @lc code=end