LeetCode //C - 102. Binary Tree Level Order Traversal

102. Binary Tree Level Order Traversal

Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level).

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[15,7]]

Example 2:

Input: root = [1]
Output: [[1]]

Example 3:

Input: root = []
Output: []

Constraints:

  • The number of nodes in the tree is in the range [0, 2000].
  • -1000 <= Node.val <= 1000

From: LeetCode

Link: 102. Binary Tree Level Order Traversal


Solution:

Ideas:
  1. Create a queue to hold the nodes of the tree.
  2. Enqueue the root node to the queue.
  3. While the queue is not empty:
  • Calculate the number of nodes at the current level.
  • Allocate memory for these nodes.
  • For each node at the current level:
    • Dequeue the node from the queue.
    • Add the node's value to the result.
    • Enqueue the node's left child (if it exists).
    • Enqueue the node's right child (if it exists).
  1. Return the result.
Code:
c 复制代码
/**
 * 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** levelOrder(struct TreeNode* root, int* returnSize, int** returnColumnSizes) {
    if (!root) {
        *returnSize = 0;
        return NULL;
    }

    // Create a queue for BFS
    struct TreeNode** queue = malloc(2000 * sizeof(struct TreeNode*));
    int front = 0, rear = 0;

    // Enqueue root
    queue[rear++] = root;

    int** result = malloc(2000 * sizeof(int*));
    *returnColumnSizes = malloc(2000 * sizeof(int));
    *returnSize = 0;

    while (front < rear) {
        // Number of nodes at current level
        int levelSize = rear - front;
        (*returnColumnSizes)[*returnSize] = levelSize;

        // Allocate memory for nodes of this level
        result[*returnSize] = malloc(levelSize * sizeof(int));

        for (int i = 0; i < levelSize; i++) {
            // Dequeue node
            struct TreeNode* current = queue[front++];

            // Add node's value to result
            result[*returnSize][i] = current->val;

            // Enqueue left child
            if (current->left) {
                queue[rear++] = current->left;
            }
            
            // Enqueue right child
            if (current->right) {
                queue[rear++] = current->right;
            }
        }
        (*returnSize)++;
    }
    free(queue);
    return result;
}
相关推荐
纪元A梦5 小时前
贪心算法应用:化工反应器调度问题详解
算法·贪心算法
阿让啊6 小时前
C语言strtol 函数使用方法
c语言·数据结构·c++·单片机·嵌入式硬件
深圳市快瞳科技有限公司6 小时前
小场景大市场:猫狗识别算法在宠物智能设备中的应用
算法·计算机视觉·宠物
liulilittle6 小时前
OPENPPP2 —— IP标准校验和算法深度剖析:从原理到SSE2优化实现
网络·c++·网络协议·tcp/ip·算法·ip·通信
superlls9 小时前
(算法 哈希表)【LeetCode 349】两个数组的交集 思路笔记自留
java·数据结构·算法
田里的水稻9 小时前
C++_队列编码实例,从末端添加对象,同时把头部的对象剔除掉,中的队列长度为设置长度NUM_OBJ
java·c++·算法
纪元A梦9 小时前
贪心算法应用:保险理赔调度问题详解
算法·贪心算法
Florence239 小时前
计算机组成原理:GPU架构、并行计算、内存层次结构等
c语言
Jayden_Ruan10 小时前
C++逆向输出一个字符串(三)
开发语言·c++·算法
不吃鱼的羊10 小时前
启动文件Startup_vle.c
c语言·开发语言