297. Serialize and Deserialize Binary Tree
# bfs(큐) 또는 dfs(preorder 순회: 나-좌-우)
# null 대신 N 으로
DFS: preorder 가 더 쉽게 구현 가능
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Codec:
def serialize(self, root):
"""Encodes a tree to a single string.
:type root: TreeNode
:rtype: str
"""
res = []
def dfs(node):
if not node:
res.append("N")
return
res.append(str(node.val))
dfs(node.left)
dfs(node.right)
dfs(root)
return ",".join(res) # 각 element 마다 , 로 구분줘서 string 으로 변경
def deserialize(self, data):
"""Decodes your encoded data to tree.
:type data: str
:rtype: TreeNode
"""
vals = data.split(",")
self.i = 0
def dfs():
if vals[self.i] == "N": # base case
self.i +=1
return None
node = TreeNode(int(vals[self.i]))
self.i +=1
node.left = dfs() # dfs 콜하면, 내부적으로 self.i 증가
node.right = dfs()
return node
return dfs()
# Your Codec object will be instantiated and called as such:
# ser = Codec()
# deser = Codec()
# ans = deser.deserialize(ser.serialize(root))
'LeetCode > 주제별 보충' 카테고리의 다른 글
Backtracking: 40. Combination Sum II (0) | 2025.01.19 |
---|---|
Backtracking: 78. Subsets (0) | 2025.01.19 |
Tree: 235. Lowest Common Ancestor of a Binary Search Tree (0) | 2025.01.16 |
Tree: 572. Subtree of Another Tree (0) | 2025.01.16 |
Tree: 110. Balanced Binary Tree (0) | 2025.01.16 |