LeetCode-in-All
102. Binary Tree Level Order Traversal
Medium
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
Solution
type TreeNode struct {
Val int
Left *TreeNode
Right *TreeNode
}
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func levelOrder(root *TreeNode) [][]int {
if root == nil {
return [][]int{}
}
queue := make([]*TreeNode, 0)
queue = append(queue, root)
ans := make([][]int, 0)
for len(queue) > 0 {
levelSize := len(queue)
temp := make([]int, 0)
for levelSize > 0 {
node := queue[0]
queue = queue[1:]
temp = append(temp, node.Val)
if node.Left != nil {
queue = append(queue, node.Left)
}
if node.Right != nil {
queue = append(queue, node.Right)
}
levelSize--
}
ans = append(ans, temp)
}
return ans
}