This approach utilizes a queue to perform a Breadth-First Search (BFS) on the binary tree. Starting from the root, traverse each level of the tree and store values of nodes in separate lists for each level. By using a queue, we can efficiently keep track of nodes at the current level and their respective children for the next level.
Time Complexity: O(n), where n is the number of nodes in the tree because each node is visited once.
Space Complexity: O(n), as we store all nodes at each level in the queue.
1function TreeNode(val, left, right) {
2 this.val = (val===undefined ? 0 : val);
3 this.left = (left===undefined ? null : left);
4 this.right = (right===undefined ? null : right);
5}
6
7function levelOrder(root) {
8 if (!root) return [];
9 const result = [];
10 const queue = [root];
11 while (queue.length > 0) {
12 const levelSize = queue.length;
13 const levelNodes = [];
14 for (let i = 0; i < levelSize; i++) {
15 const node = queue.shift();
16 levelNodes.push(node.val);
17 if (node.left) queue.push(node.left);
18 if (node.right) queue.push(node.right);
19 }
20 result.push(levelNodes);
21 }
22 return result;
23}In this implementation, we use a simple array to serve as the queue. Begin with the root node in the queue, and process all nodes level by level. For each node, enqueue its children. Capture node values into sub-arrays, which represents each level's output. Finally, return the result containing all levels.
This approach involves using a Depth-First Search (DFS) where we pass along the current level during the recursive calls. By keeping track of the current depth, we can directly add node values to their appropriate level in our result list, creating new levels in our result list as needed.
Time Complexity: O(n) since each node is visited once.
Space Complexity: O(n), considering the space needed for the result and recursive call stack.
1#include <stdio.h>
2#include <stdlib.h>
3
4struct TreeNode {
5 int val;
6 struct TreeNode *left;
7 struct TreeNode *right;
8};
9
10void dfs(struct TreeNode* node, int level, int** result, int* returnSize, int** resultColSizes) {
11 if (!node) return;
12 if (level >= *returnSize) {
13 *returnSize += 1;
14 result[*returnSize - 1] = (int*)malloc(2000 * sizeof(int));
15 (*resultColSizes)[*returnSize - 1] = 0;
16 }
17 result[level][(*resultColSizes)[level]] = node->val;
18 (*resultColSizes)[level] += 1;
19 dfs(node->left, level + 1, result, returnSize, resultColSizes);
20 dfs(node->right, level + 1, result, returnSize, resultColSizes);
21}
22
23int** levelOrder(struct TreeNode* root, int* returnSize, int** resultColSizes) {
24 *returnSize = 0;
25 int** result = (int**)malloc(2000 * sizeof(int*));
26 *resultColSizes = (int*)malloc(2000 * sizeof(int));
27 dfs(root, 0, result, returnSize, resultColSizes);
28 return result;
29}This C implementation makes use of recursive DFS calls. Starting with an empty result array, we traverse the tree. For each node, check if the current level is present in the result array. If not, allocate space for a new level's data. Store node values at their respective levels and recurse into children, incrementing the level parameter. This allows us to collect all nodes according to their levels in the tree.