LeetCode-in-Swift
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
/**
* Definition for a binary tree node.
* public class TreeNode {
* public var val: Int
* public var left: TreeNode?
* public var right: TreeNode?
* public init() { self.val = 0; self.left = nil; self.right = nil; }
* public init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; }
* public init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {
* self.val = val
* self.left = left
* self.right = right
* }
* }
*/
class Solution {
func levelOrder(_ root: TreeNode?) -> [[Int]] {
var result = [[Int]]()
guard let root = root else {
return result
}
var queue: [TreeNode?] = [root, nil]
var level = [Int]()
while !queue.isEmpty {
let node = queue.removeFirst()
if let node = node {
level.append(node.val)
if let left = node.left {
queue.append(left)
}
if let right = node.right {
queue.append(right)
}
} else {
result.append(level)
level = [Int]()
if !queue.isEmpty {
queue.append(nil)
}
}
}
return result
}
}