"""
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
"""
# 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):
        if root is None:
            return
        if root.next and root.right:
            root.right.next = root.next.left
        if root.left:
            root.left.next = root.right
        self.connect(root.left)
        self.connect(root.right)

"""
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
"""
# 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):
        if root is None:
            return
        if root.next and root.right:
            root.right.next = root.next.left
        if root.left:
            root.left.next = root.right
        self.connect(root.left)
        self.connect(root.right)

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 # 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 # @param root, a tree link node # @return nothing