116. Populating Next Right Pointers in Each Node

[LLancelot]

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.

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.

代码

"""
# Definition for a Node.
class Node(object):
    def __init__(self, val, left, right, next):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""
class Solution(object):
    def connect(self, root):
        """
        :type root: Node
        :rtype: Node
        """
        if not root or not root.left:
            return root
        root.left.next = root.right
        if root.next:
            root.right.next = root.next.left
        self.connect(root.left)
        self.connect(root.right)
        return root        

class Solution(object):
    def connect(self, root):
        """
        :type root: Node
        :rtype: Node
        """
        if not root:
            return None
        queue = deque()
        queue.append(root)

        while queue:
            preNode = Node()
            queuelen = len(queue)
            for _ in range(queuelen):
                cur = queue.popleft()
                if preNode:
                    preNode.next = cur
                preNode = cur

                if cur.left:
                    queue.append(cur.left)
                if cur.right:
                    queue.append(cur.right)

        return root