✅Super Easy (C++/Java/Python) Solution With Detailed Explanation✅
Intuition\nThe intuition of level-order traversal (BFS) approach is to efficiently finds the bottom-left value of a binary tree by systematically exploring each
# Intuition\nThe intuition of level-order traversal (BFS) approach is to efficiently finds the bottom-left value of a binary tree by systematically exploring each level and updating the candidate value with the first node of each level.\n\n# Approach\n\n1. **Initialization**: A queue is used to facilitate the level-order traversal. The root node is initially pushed into the queue.\n   \n2. **Traversal**: The process continues as long as the queue is not empty. At the beginning of each level, the size of the queue is noted, which indicates the number of nodes at that level. \n\n3. **Level Processing**: For each node in the current level, the algorithm performs the following steps:\n   - **Leftmost Value Update**: At the start of processing a new level, the value of the first node (which is the leftmost node of that level) is stored. This ensures that if this level is the last, its leftmost value is captured.\n   - **Child Nodes Enqueue**: The left and right child nodes of the current node (if present) are added to the queue. This ensures that the next level of the tree will be processed in the subsequent iteration.\n\n4. **Result**: After traversing the entire tree, the last updated value represents the bottom-left value of the tree, which is returned as the result.\n\n### Complexity Analysis\n\n- **Time Complexity**: O(N), where N is the number of nodes in the binary tree.\n  - Each node is visited exactly once during the level-order traversal. The operations performed per node (enqueuing/dequeuing) are constant time operations, leading to a linear time complexity relative to the number of nodes.\n\n- **Space Complexity**: O(W), where W is the maximum width of the tree.\n  - The space complexity is determined by the maximum number of nodes that can be stored in the queue at any point in time, which occurs at the level with the maximum width. In the worst case (a complete binary tree), the maximum width can be up to N/2 (for the last level), leading to a space complexity of O(N). However, for trees that are not complete, the width W can be significantly less than N/2, so the space complexity is more accurately described as O(W).\n\n# Code\n```C++ []\nclass Solution {\npublic:\n    int findBottomLeftValue(TreeNode* root) {\n        queue<TreeNode *> q;\n        q.push(root);\n\n        int ans = root->val;\n        while (q.size() != 0)\n        {\n            int size = q.size();\n            ans = q.front()->val;\n            for (int i=0; i<size; i++)\n            {\n                TreeNode *front = q.front();\n                q.pop();\n\n                if (front->left)\n                    q.push(front->left);\n\n                if (front->right)\n                    q.push(front->right);\n            }\n        }\n\n        return ans;\n    }\n};\n```\n```Java []\nclass Solution {\n    public int findBottomLeftValue(TreeNode root) {\n        Queue<TreeNode> q = new LinkedList<>();\n        q.offer(root);\n\n        int ans = root.val;\n        while (!q.isEmpty()) {\n            int size = q.size();\n            ans = q.peek().val; // Get the leftmost value of the current level\n            for (int i = 0; i < size; i++) {\n                TreeNode front = q.poll();\n\n                if (front.left != null)\n                    q.offer(front.left);\n\n                if (front.right != null)\n                    q.offer(front.right);\n            }\n        }\n\n        return ans;\n    }\n}\n```\n```Python []\nclass Solution:\n    def findBottomLeftValue(self, root: Optional[TreeNode]) -> int:\n        queue = deque([root])\n\n        while queue:\n            node_count = len(queue)\n            for i in range(node_count):\n                node = queue.popleft()\n                \n                # Enqueue left child first to ensure the leftmost node is processed first\n                if node.left:\n                    queue.append(node.left)\n                if node.right:\n                    queue.append(node.right)\n\n                # For the first node of the last level, this will be the answer\n                if i == 0:\n                    ans = node.val\n\n        return ans\n```
['Breadth-First Search', 'C++', 'Java', 'Python3']