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.
1from collections import deque
2
3class TreeNode:
4 def __init__(self, val=0, left=None, right=None):
5 self.val = val
6 self.left = left
7 self.right = right
8
9def levelOrder(root):
10 if not root:
11 return []
12 result = []
13 queue = deque([root])
14 while queue:
15 level_size = len(queue)
16 level_nodes = []
17 for _ in range(level_size):
18 node = queue.popleft()
19 level_nodes.append(node.val)
20 if node.left:
21 queue.append(node.left)
22 if node.right:
23 queue.append(node.right)
24 result.append(level_nodes)
25 return resultWe use a deque from the collections module to easily append and pop nodes from the queue. Starting with the root node if it is not null, we add it to the queue. For each level, determine the number of nodes at that level (i.e., the size of the queue), then iterate over them, adding each node's children to the queue for the next level. Collect and return the node values in a nested list representing each level.
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.
1using System;
2using System.Collections.Generic;
3
4public class TreeNode {
5 public int val;
6 public TreeNode left;
7 public TreeNode right;
8 public TreeNode(int x) { val = x; }
9}
10
11public class Solution {
12 public IList<IList<int>> LevelOrder(TreeNode root) {
13 var levels = new List<IList<int>>();
14 DFS(root, 0, levels);
15 return levels;
16 }
17
18 private void DFS(TreeNode node, int level, List<IList<int>> levels) {
19 if (node == null) return;
20 if (level == levels.Count)
21 levels.Add(new List<int>());
22 levels[level].Add(node.val);
23 DFS(node.left, level + 1, levels);
24 DFS(node.right, level + 1, levels);
25 }
26}This C# example employs a recursive DFS method. Starting from the root, if current depth matches the level count, a new level is added. Each node value is added to its respective level list during traversal. The function recursively processes left and then right children, ensuring nodes are placed in their correct depth-lists.