LC-0116 Medium LeetCode

116. Populating Next Right Pointers in Each Node

Linked List Tree Depth-First Search Breadth-First Search Binary Tree

Read the full problem statement on LeetCode.

Difficulty: medium Acceptance: 65% Topics: Linked List, Tree, Depth-First Search, Breadth-First Search, Binary Tree

Reading material

Linked Lists 14 min Trees & Traversals 18 min

Reference solution (spoiler · python)

# Time:  O(n)
# Space: O(1)

class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
        self.next = None

    def __repr__(self):
        if self is None:
            return "Nil"
        else:
            return "{} -> {}".format(self.val, repr(self.next))

class Solution(object):
    # @param root, a tree node
    # @return nothing
    def connect(self, root):
        head = root
        while head:
            cur = head
            while cur and cur.left:
                cur.left.next = cur.right
                if cur.next:
                    cur.right.next = cur.next.left
                cur = cur.next
            head = head.left

# Time:  O(n)
# Space: O(logn)
# recusion
class Solution2(object):
    # @param root, a tree node
    # @return nothing
    def connect(self, root):
        if root is None:
            return
        if root.left:
            root.left.next = root.right
        if root.right and root.next:
            root.right.next = root.next.left
        self.connect(root.left)
        self.connect(root.right)

Solution from kamyu104/LeetCode-Solutions · MIT