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

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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).

4. Return the result.

Code:

/**
 * 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;
}