116. Populating Next Right Pointers In Each Node - Explanation

Problem Link

Description

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.

Example 1:

Input: root = [1,2,3,4,5,6,7]

Output: [1,#,2,3,#,4,5,6,7,#]

Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.

Example 2:

Input: root = [1]

Output: [1,#]

Constraints:

  • 0 <= number of nodes in the tree <= 4095
  • -1000 <= Node.val <= 1000

Follow-up:

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

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"""
# 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: 'Optional[Node]') -> 'Optional[Node]':
        if not root:
            return None

        q = deque([root])
        while q:
            levelSize = len(q)
            while levelSize:
                node = q.popleft()
                if levelSize > 1:
                    node.next = q[0]
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
                levelSize -= 1

        return root

Time & Space Complexity

  • Time complexity: O(n)O(n)
  • Space complexity: O(log⁡n)O(\log n)
"""
# 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: 'Optional[Node]') -> 'Optional[Node]':
        mp = {}

        def dfs(node, depth):
            if not node:
                return

            if depth not in mp:
                mp[depth] = node
            else:
                mp[depth].next = node
                mp[depth] = node

            dfs(node.left, depth + 1)
            dfs(node.right, depth + 1)

        dfs(root, 0)
        return root

Time & Space Complexity

  • Time complexity: O(n)O(n)
  • Space complexity: O(log⁡n)O(\log n)

3. Depth First Search (Optimal)

"""
# 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: 'Optional[Node]') -> 'Optional[Node]':
        if not root:
            return root

        if root.left:
            root.left.next = root.right
            if root.next:
                root.right.next = root.next.left

            self.connect(root.left)
            self.connect(root.right)

        return root

Time & Space Complexity

  • Time complexity: O(n)O(n)
  • Space complexity: O(log⁡n)O(\log n) for the recursion stack.

4. Breadth First Search (Optimal)

"""
# 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: 'Optional[Node]') -> 'Optional[Node]':
        cur, nxt = root, root.left if root else None

        while cur and nxt:
            cur.left.next = cur.right
            if cur.next:
                cur.right.next = cur.next.left

            cur = cur.next
            if not cur:
                cur = nxt
                nxt = cur.left

        return root

Time & Space Complexity

  • Time complexity: O(n)O(n)
  • Space complexity: O(1)O(1)