16. Binary Tree Level Order Traversal
Pattern: BFS
Problem
Given the root of a binary tree, return values grouped by level.
Example:
3
/ \
9 20
/ \
15 7
Output: [[3], [9, 20], [15, 7]]Solution
func levelOrderTraversal(root *TreeNode) [][]int {
if root == nil {
return nil
}
result := [][]int{}
queue := []*TreeNode{root}
for len(queue) > 0 {
levelSize := len(queue)
level := []int{}
for i := 0; i < levelSize; i++ {
node := queue[0]
queue = queue[1:]
level = append(level, node.Val)
if node.Left != nil {
queue = append(queue, node.Left)
}
if node.Right != nil {
queue = append(queue, node.Right)
}
}
result = append(result, level)
}
return result
}Snapshot levelSize before processing — this separates levels cleanly.