LeetCode-in-Net.github.io
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
using LeetCodeNet.Com_github_leetcode;
/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
public class Solution {
public IList<IList<int>> LevelOrder(TreeNode root) {
IList<IList<int>> result = new List<IList<int>>();
if (root == null) {
return result;
}
Queue<TreeNode> queue = new Queue<TreeNode>();
queue.Enqueue(root);
queue.Enqueue(null);
List<int> level = new List<int>();
while (queue.Count > 0) {
root = queue.Dequeue();
while (queue.Count > 0 && root != null) {
level.Add((int)root.val);
if (root.left != null) {
queue.Enqueue(root.left);
}
if (root.right != null) {
queue.Enqueue(root.right);
}
root = queue.Dequeue();
}
result.Add(level);
level = new List<int>();
if (queue.Count > 0) {
queue.Enqueue(null);
}
}
return result;
}
}