[Tree/LinkList]117. Populating Next Right Pointers in Each Node II

• 分类：Tree/List
• 时间复杂度: O(n) 相当于把所有节点都遍历一遍
• 空间复杂度: O(1)

117. Populating Next Right Pointers in Each Node II

Given a binary tree

struct Node {

 int val;

 Node *left;

 Node *right;

 Node *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.

Example:

image

Input: {"$id":"1","left":{"$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":{"$id":"5","left":null,"next":null,"right":{"$id":"6","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}

Output: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":null,"right":null,"val":7},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"6","left":null,"next":null,"right":{"$ref":"5"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"6"},"val":1}

Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B.

Note:

• You may only use constant extra space.

• Recursive approach is fine, implicit stack space does not count as extra space for this problem.

代码：

GTH代码思路：

"""
# Definition for a Node.
class Node:
    def __init__(self, val, left, right, next):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""
class Solution:
    
    def connect(self, root: 'Node') -> 'Node':
        
        if root==None:
            return None
        
        res=Node(0,None,None,None)
        res.left=root
        
        while root!=None:
            dummy=Node(0,None,None,None)
            curr=dummy
            while root!=None:
                if root.left!=None:
                    curr.next=root.left
                    curr=curr.next
                if root.right!=None:
                    curr.next=root.right
                    curr=curr.next
                root=root.next
            curr.next=None
            root=dummy.next
        
        return res.left

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讨论：

1.自己写了一个代码怎么搞都通不过= 。=
2.最后看了GTH的思路，每次都觉得GTH思路好清晰好棒
3.这个题好像有点像linklist和tree的结合
4.每次都搞一条新的链条