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
醉颜凉34 分钟前
计算(a+b)/c的值
java·c语言·数据结构·c++·算法
liyinuo20171 小时前
如何使用GCC手动编译stm32程序
c语言·arm开发·stm32·单片机·嵌入式硬件
武昌库里写JAVA2 小时前
SpringCloud+SpringCloudAlibaba学习笔记
java·开发语言·算法·spring·log4j
小咖拉眯2 小时前
第十六届蓝桥杯模拟赛第二期题解—Java
java·数据结构·算法·蓝桥杯·图搜索算法
Sunyanhui12 小时前
力扣 最长回文字串-5
算法·leetcode·职场和发展
csdn_aspnet2 小时前
C# 程序来计算三角形的面积(Program to find area of a triangle)
算法·c#
xiangxiang-2 小时前
目标检测,图像分割,超分辨率重建
算法·机器学习·支持向量机
一直学习永不止步2 小时前
LeetCode题练习与总结:数组中两个数的最大异或值--421
java·算法·leetcode·字典树·数组·位运算·哈希表
机器学习之心2 小时前
异常检测 | 高斯分布拟合算法异常数据检测(Matlab)
算法·数学建模·matlab·异常数据检测
Ning_.2 小时前
力扣第 66 题 “加一”
算法·leetcode