LeetCode 150. 彩灯装饰记录 II
原题链接
简单
作者:
Noe1017
,
2023-10-05 11:02:38
,
所有人可见
,
阅读 79
传送门
思路:
- 利用$bfs$,是层序遍历这一性质,我们每次只需要将对应层数大小的节点都放到数组中保存起来即可
C++
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<vector<int>> decorateRecord(TreeNode* root) {
vector<vector<int>>ans;
if(root==nullptr)return ans;
queue<TreeNode*>q;
q.push(root);
while(!q.empty())
{
vector<int>tmp;
int len=(int)q.size();
for(int i=1;i<=len;i++)
{
auto t=q.front();q.pop();
tmp.push_back(t->val);
if(t->left)q.push(t->left);
if(t->right)q.push(t->right);
}
ans.push_back(tmp);
}
return ans;
}
};
Python3
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def decorateRecord(self, root: Optional[TreeNode]) -> List[List[int]]:
ans = []
if not root:return ans
res = []
q = collections.deque()
q.append(root)
while q :
tmp = []
for _ in range(len(q)):
t = q.popleft()
tmp.append(t.val)
if t.left:q.append(t.left)
if t.right:q.append(t.right)
res.append(tmp)
return res
哇,限时返场了耶
哈哈