Merge pull request #131 from sounmind/main

Author: sounmind

binary-tree-level-order-traversal/evan.js

@@ -0,0 +1,37 @@
+/**
+ * @param {TreeNode} root
+ * @return {number[][]}
+ */
+var levelOrder = function (root) {
+  const result = [];
+
+  const dfs = (node, level) => {
+    if (!node) {
+      return;
+    }
+
+    if (!result[level]) {
+      result[level] = [];
+    }
+
+    result[level].push(node.val);
+
+    dfs(node.left, level + 1);
+    dfs(node.right, level + 1);
+  };
+
+  dfs(root, 0);
+
+  return result;
+};
+
+/**
+ * Time Complexity: O(n)
+ * - The function visits each node exactly once, making it O(n).
+ * - Here, n is the number of nodes in the binary tree.
+ *
+ * Space Complexity: O(n)
+ * - The space complexity is determined by the recursion stack.
+ * - In the worst case, the recursion stack could store all nodes in the binary tree.
+ * - Thus, the space complexity is O(n).
+ */

lowest-common-ancestor-of-a-binary-search-tree/evan.js

@@ -0,0 +1,29 @@
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val) {
+ *     this.val = val;
+ *     this.left = this.right = null;
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @param {TreeNode} p
+ * @param {TreeNode} q
+ * @return {TreeNode}
+ */
+var lowestCommonAncestor = function (root, p, q) {
+  if (root.val < p.val && root.val < q.val) {
+    return lowestCommonAncestor(root.right, p, q);
+  }
+
+  if (root.val > p.val && root.val > q.val) {
+    return lowestCommonAncestor(root.left, p, q);
+  }
+
+  return root;
+};
+
+/**
+ * Time Complexity: O(h), where h is the height of the binary search tree.
+ * Space Complexity: O(h), where h is the height of the binary search tree.
+ */

remove-nth-node-from-end-of-list/evan.js

@@ -0,0 +1,33 @@
+/**
+ * @param {ListNode} head
+ * @param {number} n
+ * @return {ListNode}
+ */
+function removeNthFromEnd(head, n) {
+  let dummy = new ListNode(0);
+  dummy.next = head;
+  let first = dummy;
+  let second = dummy;
+
+  // Move the first pointer n+1 steps ahead
+  for (let i = 0; i <= n; i++) {
+    first = first.next;
+  }
+
+  // Move both pointers until the first pointer reaches the end
+  // Second pointer will be at the (n+1)th node from the end
+  while (first !== null) {
+    first = first.next;
+    second = second.next;
+  }
+
+  // Remove the nth node from the end
+  second.next = second.next.next;
+
+  return dummy.next;
+}
+
+/**
+ * Time Complexity: O(n)
+ * Space Complexity: O(1)
+ */

reorder-list/evan.js

@@ -0,0 +1,77 @@
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @return {void} Do not return anything, modify head in-place instead.
+ */
+var reorderList = function (head) {
+  const middleNode = findMiddleNode(head);
+  let reversedHalf = reverseList(middleNode.next);
+  middleNode.next = null;
+
+  let half = head;
+
+  while (reversedHalf) {
+    const nextForward = half.next;
+    const nextBackward = reversedHalf.next;
+
+    half.next = reversedHalf;
+    reversedHalf.next = nextForward;
+
+    reversedHalf = nextBackward;
+    half = nextForward;
+  }
+};
+/**
+ * Time Complexity: O(n)
+ * - Finding the middle node takes O(n).
+ * - Reversing the second half takes O(n).
+ * - Merging the two halves takes O(n).
+ * - Overall, the time complexity is O(n) + O(n) + O(n) = O(n).
+ *
+ * Space Complexity: O(1)
+ * - We only use a constant amount of extra space (pointers),
+ *   so the space complexity is O(1).
+ */
+
+/**
+ * @param {ListNode} head
+ * @return {ListNode}
+ */
+var reverseList = function (head) {
+  let currentNode = head;
+  let previousNode = null;
+
+  while (currentNode !== null) {
+    const nextNode = currentNode.next;
+
+    currentNode.next = previousNode;
+    previousNode = currentNode;
+    currentNode = nextNode;
+  }
+
+  return previousNode;
+};
+
+/**
+ * @param {ListNode} head
+ * @return {ListNode}
+ */
+var findMiddleNode = function (head) {
+  if (!head) return null;
+
+  let slow = head;
+  let fast = head;
+
+  while (fast !== null && fast.next !== null) {
+    slow = slow.next;
+    fast = fast.next.next;
+  }
+
+  return slow;
+};

validate-binary-search-tree/evan.js

@@ -0,0 +1,28 @@
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @return {boolean}
+ */
+function isValidBST(root, low = -Infinity, high = Infinity) {
+  if (root === null) {
+    return true;
+  }
+
+  if (root.val <= low || root.val >= high) {
+    return false;
+  }
+
+  return (
+    // left root should be less than current root
+    isValidBST(root.left, low, root.val) &&
+    // right root should be greater than current root
+    isValidBST(root.right, root.val, high)
+  );
+}