Binary Tree Explorer: Mastered JavaScript | 100.00% Efficiency Seeker
Intuition\nThe code aims to find and return the leftmost value in the last level of a binary tree. The use of a deque and a level-order traversal suggests a bre
# Intuition\nThe code aims to find and return the leftmost value in the last level of a binary tree. The use of a deque and a level-order traversal suggests a breadth-first search (BFS) approach to efficiently traverse the tree level by level.\n\n# Approach\n1.Initialize a deque queue with the root node and a variable leftmost_value to None.\n2.Perform a while loop while the queue is not empty:\n   Dequeue a node from the left of the deque.\n   Update leftmost_value with the value of the dequeued node.\n   Enqueue the right child if it exists.\n   Enqueue the left child if it exists.\n3.Continue the process until all nodes at the last level are processed.\n4.Return the leftmost_value.\n\n# Complexity\n  Time complexity: O(n)O(n)O(n), where nnn is the number of nodes in the binary tree. Each node is processed once during the level-order traversal.\n  Space complexity: O(m)O(m)O(m), where mmm is the maximum number of nodes at any level in the binary tree. In the worst case, the queue would store all nodes at the maximum level.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n/**\n * @param {TreeNode} root\n * @return {number}\n */\nvar findBottomLeftValue = function(root) {\n\n    const queue = [root];\n    let leftmostValue;\n\n    while (queue.length > 0) {\n        const node = queue.shift();\n\n        leftmostValue = node.val;\n\n        if (node.right) {\n            queue.push(node.right);\n        }\n        if (node.left) {\n            queue.push(node.left);\n        }\n    }\n\n    return leftmostValue;\n};\n```
['JavaScript']