LeetCode 0102. Binary Tree Level Order Traversal二叉樹的層次遍歷【Medium】【Python】【BFS】
Problem
Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).
For example:
Given binary tree [3,9,20,null,null,15,7]
,
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
問題
給定一個二叉樹,返回其按層次遍歷的節點值。 (即逐層地,從左到右訪問所有節點)。
例如:
給定二叉樹: [3,9,20,null,null,15,7]
,
3
/ \
9 20
/ \
15 7
返回其層次遍歷結果:
[
[3],
[9,20],
[15,7]
]
思路
BFS
解法一:非遞歸
當隊列不爲空:
當前層打印循環:
隊首元素出隊,記爲 node
將 node.val 添加到 temp 尾部
若左(右)子節點不爲空,則將左(右)子節點加入隊列
把當前 temp 中的所有元素加入 res
時間複雜度: O(n),n 爲二叉樹的節點數。
空間複雜度: O(n),n 爲二叉樹的節點數。
Python3代碼
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
# solution one: 非遞歸
import collections
if not root:
return []
res, q = [], collections.deque()
q.append(root)
while q:
# 輸出是二維數組
temp = []
for x in range(len(q)):
node = q.popleft()
temp.append(node.val)
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
res.append(temp)
return res
解法二:遞歸
Python3代碼
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
# solution two: 遞歸
res = []
def helper(root, depth):
if not root:
return
if len(res) == depth:
res.append([])
res[depth].append(root.val)
helper(root.left, depth + 1)
helper(root.right, depth + 1)
helper(root , 0)
return res