Java | BFS | 10 lines | Clean code
Complexity\n- Time complexity: O(n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(n)\n Add your space complexity here, e.g. O(n) \n\n# Co
# Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n  public int findBottomLeftValue(TreeNode root) {\n    var queue = new ArrayDeque<TreeNode>();\n    queue.offer(root);\n    var node = root;\n\n    while (!queue.isEmpty()) {\n      node = queue.poll();\n\n      if (node.right != null) queue.offer(node.right);\n      if (node.left != null) queue.offer(node.left);\n    }\n    return node.val;\n  }\n}\n```\nIf you like my solution, please upvote it!
['Breadth-First Search', 'Binary Tree', 'Java']