117. Populating Next Right Pointers in Each Node II

https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/#/description

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

• You may only use constant extra space.

For example,
Given the following binary tree,

         1
       /  
      2    3
     /     
    4   5    7

After calling your function, the tree should look like:

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

 

 

Sol 1:

 

Recursion.

 

# 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):
        
        
    # recursion
    # it can solve Populating Next Right Pointers in Each Node I.
    # To handle the right sub node, the first thought pops out is BFS.  However, BFS does not take up constant space
    # Time O(n) Space(1)
    
        if not root:
            return
        
        # dummy stores node on the previous level. 
        dummy = TreeLinkNode(0)
        curr = root
        prev = dummy
        while curr:

            if curr.left:
                prev.next = curr.left
                prev = prev.next
            if curr.right:
                prev.next = curr.right
                prev = prev.next   
            curr = curr.next
        
        self.connect(dummy.next)
    

 

 Note:

1 Use built in class given by the question. 

Get the first element of a tree using class TreeLinkNode.

dummy = TreeLinkNode(0)

Sol 2:

Iteration. 

Not very different from the recursion method. Just change the way we forward the dummy variable.

"

 The algorithm is a BFS or level order traversal. We go through the tree level by level. node is the pointer in the parent level, tail is the tail pointer in the child level.
The parent level can be view as a singly linked list or queue, which we can traversal easily with a pointer.
Connect the tail with every one of the possible nodes in child level, update it only if the connected node is not nil.
Do this one level by one level. The whole thing is quite straightforward.

"

# 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, node):

    # iteration
        tail = dummy = TreeLinkNode(0)
        while node:
            tail.next = node.left
            if tail.next:
                tail = tail.next
            tail.next = node.right
            if tail.next:
                tail = tail.next
            node = node.next
            if not node:
                tail = dummy
                node = dummy.next