🌟 JavaScript || Professional Approach and In-Depth Explanation
Intuition\nWe need to find the value of the leftmost node at the deepest level in a binary tree. This suggests a traversal of the tree to find the depth and the
# Intuition\nWe need to find the value of the leftmost node at the deepest level in a binary tree. This suggests a traversal of the tree to find the depth and the leftmost node at that depth.\n\n# Approach\nWe\'ll use a Depth-First Search (DFS) traversal technique to explore the tree. During the traversal, we\'ll keep track of both the depth and the leftmost value encountered so far. Whenever we encounter a node deeper than the previously recorded depth, we update the leftmost value with the current node\'s value. We traverse the left subtree first, then the right subtree, ensuring that we always explore the leftmost nodes at each level before moving to the right.\n\n# Complexity\n- Time complexity: \n    - Traversing through each node of the tree takes O(n) time, where n is the number of nodes in the tree.\n- Space complexity:\n    - The space complexity is O(h), where h is the height of the tree. In the worst-case scenario, where the tree is skewed, the space complexity could be O(n). This space is used for the call stack during the recursive DFS traversal.\n\n# Code\n```javascript\n/**\n * Definition for a binary tree node.\n * function TreeNode(val, left, right) {\n *     this.val = (val===undefined ? 0 : val)\n *     this.left = (left===undefined ? null : left)\n *     this.right = (right===undefined ? null : right)\n * }\n */\n\n/**\n * @param {TreeNode} root\n * @return {number}\n */\nvar findBottomLeftValue = function(root) {\n    // Variable to store the bottom-left value\n    let leftmostValue;\n    // Variable to track the maximum depth\n    let maxDepth = -1;\n\n    // DFS function to traverse the tree\n    const dfs = (node, depth) => {\n        // If current node is null, return\n        if (!node) return;\n\n        // If current depth is greater than maxDepth,\n        // update leftmostValue and maxDepth\n        if (depth > maxDepth) {\n            leftmostValue = node.val;\n            maxDepth = depth;\n        }\n\n        // Recur for left and right children with increased depth\n        dfs(node.left, depth + 1);\n        dfs(node.right, depth + 1);\n    };\n\n    // Start DFS from root with depth 0\n    dfs(root, 0);\n\n    // Return the leftmostValue found during DFS\n    return leftmostValue;\n};\n```\n\nThis code utilizes a Depth-First Search approach to efficiently find the bottom-left value of the binary tree.
['JavaScript']