[leetcode] 116. Populating Next Right Pointers in Each Node @ python

原题

Given a binary tree

struct TreeLinkNode {
TreeLinkNode *left;
TreeLinkNode *right;
TreeLinkNode *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.

Note:

You may only use constant extra space.
Recursive approach is fine, implicit stack space does not count as extra space for this problem.
You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
Example:

Given the following perfect binary tree,

 1

/
2 3
/ \ /
4 5 6 7
After calling your function, the tree should look like:

 1 -> NULL

/
2 -> 3 -> NULL
/ \ /
4->5->6->7 -> NULL

解法1

BFS, 按层遍历, 将每层的节点加到q, 然后对q遍历, 将每个节点指向下一个节点, 最后一个节点指向None.
Time: O(n)
Space: O(1)

代码

# Definition for binary tree with next pointer.
# class TreeLinkNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#         self.next = None

class Solution:
    # @param root, a tree link node
    # @return nothing
    def connect(self, root):
        if not root:
            return
        q = [root]
        while q:
            for i in range(1, len(q)):
                q[i-1].next = q[i]
            q[-1].next = None
            new_q = []
            for node in q:
                if node.left:
                    new_q.append(node.left)
                if node.right:
                    new_q.append(node.right)
            q = new_q

解法2

BFS, 简化版

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
        """
        # base case
        if not root: return None
        q = [root]
        while q:
            for i in range(1, len(q)):
                q[i-1].next = q[i]
                
            q = [kid for node in q for kid in (node.left, node.right) if kid]
            
        return root

解法3

DFS

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
        """
        # base case
        if not root: return
        if root.right:
            root.left.next = root.right
            if root.next:
                root.right.next = root.next.left
                
        self.connect(root.left)
        self.connect(root.right)
        return root