leetcode 116. populating-next-right-pointers-in-each-node 填充每个节点的下一个右侧节点指针 python3

时间：2020-8-20

题目地址：https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node/

题目难度：Medium

题目描述：

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。

---

思路1：迭代

这个就是层级遍历，串联节点就行了

这个仿佛不符合题意，题目要求O（n）

代码段1：通过

"""
# 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') -> 'Node':
        if not root: return root
        stack = [root]
        while stack:
            n = len(stack)
            node = stack[0]
            for i in range(1, n):
                node.next = stack[i]
                node = stack[i]
            for _ in range(n):
                node = stack.pop(0)
                if node.left:
                    stack.append(node.left)
                if node.right:
                    stack.append(node.right)
        return root

总结：

1. 写不出来

---

思路2：迭代

连接的方式有两种:
第一种是这两个串联的节点都有一个共同的父节点，通过父节点就可以将这两个子节点串联起来

第二种是这两个串联的节点的父节点不同，对于这种情况，如果我们能将这一层的上一层串联好。那么可以通过父节点的next找到邻居，完成串联。

代码段2：通过

"""
# 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') -> 'Node':
        if not root: return root
        pre = root
        while pre.left:
            temp = pre
            while temp:
                temp.left.next = temp.right
                if temp.next:
                    temp.right.next = temp.next.left
                temp = temp.next
            pre = pre.left
        return root

总结：

1. 从崇礼回来还是晕晕的，下周再看吧

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后续优化：递归https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node/solution/dong-hua-yan-shi-san-chong-shi-xian-116-tian-chong/