Given a binary tree

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:

Input: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":null,"next":null,"right":{"$id":"6","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}

Output: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":null,"right":null,"val":7},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"6","left":null,"next":null,"right":{"$ref":"5"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"6"},"val":1}

Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B.

Note:

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

Solutions

1. BFS-迭代

# Time Complexity: O(n)
# Space Complexity: O(logn)
"""
# Definition for a Node.
class Node:
    def __init__(self, val, left, right, next):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""
class Solution:
    def connect(self, root: 'Node') -> 'Node':
        if not root:
            return None
        
        queue = collections.deque([root])
        
        while queue:
            pre = None
            size = len(queue)
            
            for i in range(size):
                node = queue.popleft()
                if not pre:
                    pre = node
                else:
                    pre.next = node
                    pre = node
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            
        return root
# Runtime: 388 ms, faster than 82.64% of Python3 online submissions for Populating Next Right Pointers in Each Node II.
# Memory Usage: 49.6 MB, less than 8.33% of Python3 online submissions for Populating Next Right Pointers in Each Node II.

References

  1. 117. Populating Next Right Pointers in Each Node II
  2. 相关题目
    1. 116. Populating Next Right Pointers in Each Node