116. Populating Next Right Pointers in Each Node

You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

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.

Follow up:

• You may only use constant extra space.
• Recursive approach is fine, you may assume implicit stack space does not count as extra space for this problem.

My Answer 1: Accepted (Runtime: 60 ms - 76.25% / Memory Usage: 15.8 MB - 49.21%)

"""
# Definition for a Node.
class Node:
    def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""

class Solution:
    def connect(self, root: 'Node') -> 'Node':
        if root is None:
            return root
        
        def _level_order_traversal(root):
            count = 1
            queue = [root]
            isRight = 1
            while queue:
                root = queue.pop(0)
                if queue:
                    root.next = queue[0]
                if isRight == 1 or count == isRight:
                    isRight = count * 2 + 1
                    root.next = None
                if root is not None:
                    if root.left:
                        queue.append(root.left)
                    if root.right:
                        queue.append(root.right)
                count += 1
                
        _level_order_traversal(root)
                
        return root

level order 순회를 사용해서 풀었다.
count 로 노드의 개수를 세고 isRight 로 가장 오른쪽의 노드번호를 계산해줬다.
isRight => 1, 3(1*2+1), 7(3*2+1), 15(7*2+1), 31(15*2+1), ... (count*2+1)

Recursion

Solution 1: Runtime: 68 ms - 27.12% / Memory Usage: 15.8 MB - 23.90%

"""
# Definition for a Node.
class Node:
    def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""

class Solution:
    def connect(self, root: 'Node') -> 'Node':
        ## RC ##
        ## APPROACH : RECURSION ##
        ## LOGIC ##
        ## If you keenly observe, every right nodes next pointer is LEFT child of its PARENT's NEXT NODE ##
        
		## TIME COMPLEXITY : O(N) ##
		## SPACE COMPLEXITY : O(N) ##

        def dfs(node):
            if not node : return
            if node.right and node.next:  
                node.right.next = node.next.left    # remember its a complete binary tree, no need to check if node.next.left exists or not
            if node.left:
                node.left.next = node.right
            dfs(node.left)
            dfs(node.right)
        dfs(root)
                
        return root

재귀 이용한 방식