102. Binary Tree Level Order Traversal
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
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> lists = new ArrayList<>();
Queue<TreeNode> queue = new LinkedList();
if(root==null){
return lists;
}
queue.offer(root);
while(!queue.isEmpty()){
List<Integer> list = new ArrayList<>();
int size = queue.size();
for(int i=0;i<size;i++){
TreeNode node = queue.poll();
if(node.left!=null){
queue.offer(node.left);
}
if(node.right!=null){
queue.offer(node.right);
}
list.add(node.val);
}
lists.add(list);
}
return lists;
}
}