Leetcode | 116. Populating Next Right Pointers in Each Node 二叉树增加next指针指向右边节点

本文链接：https://blog.youkuaiyun.com/zhang15953709913/article/details/84946589

本文介绍了一种在完美二叉树中填充next指针的方法，使得每个节点的next指针指向其右侧的下一个节点。文章详细解释了如何在不使用额外空间的情况下，通过迭代方式完成这一操作，同时提供了具体的Python代码实现。

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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.
Recursive approach is fine, implicit stack space does not count as extra space for this problem.
You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

Example:

Given the following perfect binary tree,

After calling your function, the tree should look like:

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

解题思路：对每个节点而言，它的左节点的next指向它的右节点，它的右节点的next指向它的next的左节点

# Definition for binary tree with next pointer.
# class TreeLinkNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#         self.next = None

class Solution:
    # @param root, a tree link node
    # @return nothing
    def connect(self, root):
        
        node = root
        while node:
            cur = node 
            while cur: # cur遍历这一层
                if cur.left and cur.right:
                    cur.left.next = cur.right
                if cur.right and cur.next and cur.next.left:
                    cur.right.next = cur.next.left
                cur = cur.next
            node = node.left # 切换到下一层