题目

LCR 150. 彩灯装饰记录 II - 力扣（LeetCode）

Python

# 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 ansqu=[root]while qu:le=len(qu)row=[]for _ in range(le):cur=qu.pop(0)row.append(cur.val)if cur.left:qu.append(cur.left)if cur.right:qu.append(cur.right)ans.append(row)return ans

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) {}* };*/#include <queue>
class Solution {
public:vector<vector<int>> decorateRecord(TreeNode* root) {vector<vector<int>> ans;queue<TreeNode*> qu;if(root!=NULL)  qu.push(root);while(!qu.empty()){int le=qu.size();vector<int> row;for(int _=0;_<le;_++){TreeNode* cur=qu.front();qu.pop();row.push_back(cur->val);if(cur->left) qu.push(cur->left);if(cur->right) qu.push(cur->right);}ans.push_back(row);}return ans;}
};

C语言

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     struct TreeNode *left;*     struct TreeNode *right;* };*/
/*** Return an array of arrays of size *returnSize.* The sizes of the arrays are returned as *returnColumnSizes array.* Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().*/
int** decorateRecord(struct TreeNode* root, int* returnSize,int** returnColumnSizes) 
{if (root == NULL) {*returnSize = 0;return root;}struct TreeNode* qu[1100];struct TreeNode* cur;int front = 0, rear = 0, row = 0, col = 0, next_row_count = 0, cur_row_count = 1;int** ans = (int**)malloc(sizeof(int*) * 1100);*returnColumnSizes = (int*)malloc(sizeof(int) * 1100);qu[rear++] = root;  //入队ans[row] = (int*)malloc(sizeof(int) * cur_row_count);(*returnColumnSizes)[row] = cur_row_count;while (front < rear) {cur = qu[front++];  //出队ans[row][col++] = cur->val;if (cur->left) {qu[rear++] = cur->left;next_row_count++;}if (cur->right) {qu[rear++] = cur->right;next_row_count++;}cur_row_count--;if (cur_row_count == 0) {cur_row_count = next_row_count;next_row_count = 0;col = 0;row++;ans[row] = (int*)malloc(sizeof(int) * cur_row_count);(*returnColumnSizes)[row] = cur_row_count;}}*returnSize = row;return ans;
}