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:
[0, 2000].-1000 <= Node.val <= 1000
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> ans = new ArrayList<>();
List<Integer> currAns = new ArrayList<>();
if(root==null){
return ans ;
}
Queue<TreeNode> main = new ArrayDeque<>();
Queue<TreeNode> helper= new ArrayDeque<>();
main.add(root);
while(main.size()>0){
TreeNode currNode = main.remove();
currAns.add(currNode.val);
if(currNode.left!=null){
helper.add(currNode.left);
}
if(currNode.right!=null){
helper.add(currNode.right);
}
if(main.size()==0){
ans.add(currAns);
currAns = new ArrayList<>();
main = helper ;
helper = new ArrayDeque<>();
}
}
return ans ;
}
}
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