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
import com_github_leetcode.TreeNode
import java.util.ArrayList
import java.util.LinkedList
import java.util.Queue
/*
* Example:
* var ti = TreeNode(5)
* var v = ti.`val`
* Definition for a binary tree node.
* class TreeNode(var `val`: Int) {
* var left: TreeNode? = null
* var right: TreeNode? = null
* }
*/
class Solution {
fun levelOrder(root: TreeNode?): List<List<Int>> {
var localRoot: TreeNode? = root
val result: MutableList<List<Int>> = ArrayList()
if (localRoot == null) {
return result
}
val queue: Queue<TreeNode> = LinkedList()
queue.add(localRoot)
queue.add(null)
var level: MutableList<Int> = ArrayList()
while (queue.isNotEmpty()) {
localRoot = queue.remove()
while (queue.isNotEmpty() && localRoot != null) {
level.add(localRoot.`val`)
if (localRoot.left != null) {
queue.add(localRoot.left)
}
if (localRoot.right != null) {
queue.add(localRoot.right)
}
localRoot = queue.remove()
}
result.add(level)
level = ArrayList()
if (queue.isNotEmpty()) {
queue.add(null)
}
}
return result
}
}