102_Binary_tree_level_order_traversal

// https://leetcode.com/problems/binary-tree-level-order-traversal/

/**
 * 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) {
        if (root == null)
            return new ArrayList<List<Integer>>();
        List<List<Integer>> traversal = new ArrayList<List<Integer>>();
        Queue<TreeNode> queue = new LinkedList();
        queue.add(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> level = new ArrayList();
            while (size > 0) {
                TreeNode node = queue.poll();
                level.add(node.val);
                if (node.left != null)
                    queue.add(node.left);
                if (node.right != null)
                    queue.add(node.right);
                size--;
            }
            traversal.add(level);
        }
        return traversal;
    }
}

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        if (root == None):
            return []
        ans = [[root.val]]
        queue = [root]
        while (len(queue) > 0):
            size = len(queue)
            while (size > 0):
                root = queue.pop(0)
                if (root.left != None):
                    queue.append(root.left)
                if (root.right != None):
                    queue.append(root.right)
                size -= 1
            ans.append([x.val for x in queue])
        return ans[:-1]