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| 1 | +// 🚀 Optimized Approach: Inorder Traversal (BST Property) |
| 2 | +// ✅ Time Complexity: O(n), n represents the number of nodes |
| 3 | +// ✅ Space Complexity: O(n), due to recursion stack (in the worst case, e.g., unbalanced tree) |
| 4 | + |
| 5 | +/** |
| 6 | + * Definition for a binary tree node. |
| 7 | + * function TreeNode(val, left, right) { |
| 8 | + * this.val = (val===undefined ? 0 : val) |
| 9 | + * this.left = (left===undefined ? null : left) |
| 10 | + * this.right = (right===undefined ? null : right) |
| 11 | + * } |
| 12 | + */ |
| 13 | +/** |
| 14 | + * @param {TreeNode} root |
| 15 | + * @param {number} k |
| 16 | + * @return {number} |
| 17 | + */ |
| 18 | +var kthSmallest = function (root, k) { |
| 19 | + // Correct Order in BST Inorder Traversal: 🌿 (left) → 🌟 (current) → 🌳 (right) |
| 20 | + // 🌿 Go Left (visit the smallest values first). |
| 21 | + // 🌟 Visit Current Node (increment count++). |
| 22 | + // 🌳 Go Right (visit larger values). |
| 23 | + |
| 24 | + let count = 0; |
| 25 | + let result = null; |
| 26 | + |
| 27 | + const inorder = (node) => { |
| 28 | + if (!node || result !== null) return; // If node is null or result is found, stop recursion |
| 29 | + |
| 30 | + inorder(node.left); |
| 31 | + |
| 32 | + count += 1; // Visit current node |
| 33 | + if (count === k) { |
| 34 | + result = node.val; |
| 35 | + return; |
| 36 | + } |
| 37 | + |
| 38 | + inorder(node.right); |
| 39 | + }; |
| 40 | + |
| 41 | + inorder(root); |
| 42 | + |
| 43 | + return result; |
| 44 | +}; |
| 45 | + |
| 46 | + |
| 47 | +// 😊 My initial approach |
| 48 | +// 🐌This is not optimal for a BST, but it works. |
| 49 | +// ✅ Time Complexity: O(n * logn), n represents the number of nodes |
| 50 | +// ✅ Space Complexity: O(n) |
| 51 | + |
| 52 | +/** |
| 53 | + * Definition for a binary tree node. |
| 54 | + * function TreeNode(val, left, right) { |
| 55 | + * this.val = (val===undefined ? 0 : val) |
| 56 | + * this.left = (left===undefined ? null : left) |
| 57 | + * this.right = (right===undefined ? null : right) |
| 58 | + * } |
| 59 | + */ |
| 60 | +/** |
| 61 | + * @param {TreeNode} root |
| 62 | + * @param {number} k |
| 63 | + * @return {number} |
| 64 | + */ |
| 65 | +// var kthSmallest = function (root, k) { |
| 66 | +// let arr = []; |
| 67 | +// const dfs = (node) => { |
| 68 | +// if (!node) return null; |
| 69 | + |
| 70 | +// if (node.val !== undefined) { |
| 71 | +// arr.push(node.val); |
| 72 | +// dfs(node.left); |
| 73 | +// dfs(node.right); |
| 74 | +// } |
| 75 | +// }; |
| 76 | + |
| 77 | +// dfs(root); |
| 78 | + |
| 79 | +// arr.sort((a, b) => a - b); |
| 80 | + |
| 81 | +// return arr[k - 1]; |
| 82 | +// }; |
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