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[leetcode]116. Populating Next Right Pointers in Each Node 본문
Study/Algorithm
[leetcode]116. Populating Next Right Pointers in Each Node
DoRoMii 2019. 8. 9. 12:31728x90
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.
"""
# 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):
parents =[]
currents = []
if root :
currents.append(root)
while currents:
parents = currents
currents = []
for p in parents:
if p.left :
currents.append(p.left)
if p.right:
currents.append(p.right)
len_p = len(parents)
for i in range(len_p):
if i < len_p-1 :
parents[i].next = parents[i+1]
else :
parents[i].next = None
return root
같은 알고리즘 더 간단한 코드
"""
# 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):
nodes = [root] if root else []
while nodes:
next_nodes = []
last = None
for node in nodes:
if last :
last.next = node
if node.left :
next_nodes.append(node.left)
if node.right:
next_nodes.append(node.right)
last = node
nodes = next_nodes
return root
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/
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