feat: Upload lowest-common-ancestor-of-a-binary-search-tree (typescript)

Author: mike2ox

lowest-common-ancestor-of-a-binary-search-tree/mike2ox.ts

@@ -0,0 +1,69 @@
+/**
+ * Source: https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree
+ * Solution: DFS를 이용해서 두 노드까지의 경로를 구해 순차적으로 비교하면서 가장 마지막 동일 노드를 추출
+ *
+ * 시간복잡도: O(N) - 최악인 경우, 전체 노드수 탐색
+ * 공간복잡도: O(N) - 최악인 경우, 전체 노드수 보관
+ *
+ * 다른 풀이
+ * - 재귀로도 해결할 것으로 보이지만 바로 구현체가 떠오르지 않음
+ */
+
+/**
+ * Definition for a binary tree node.
+ * class TreeNode {
+ *     val: number
+ *     left: TreeNode | null
+ *     right: TreeNode | null
+ *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.left = (left===undefined ? null : left)
+ *         this.right = (right===undefined ? null : right)
+ *     }
+ * }
+ */
+
+function lowestCommonAncestor(
+  root: TreeNode | null,
+  p: TreeNode | null,
+  q: TreeNode | null
+): TreeNode | null {
+  if (!root || !p || !q) return null;
+  let stack = new Array(root);
+  let pRoute: Array<TreeNode> | null;
+  let qRoute: Array<TreeNode> | null;
+  let answer: TreeNode | null;
+  const visited = new Set();
+
+  while (stack.length) {
+    const left = stack.at(-1).left;
+    if (left && !visited.has(left.val)) {
+      stack.push(left);
+      continue;
+    }
+    const right = stack.at(-1).right;
+    if (right && !visited.has(right.val)) {
+      stack.push(right);
+      continue;
+    }
+    const now = stack.pop();
+    visited.add(now.val);
+    if (now.val === q.val) {
+      qRoute = [...stack, now];
+      continue;
+    }
+    if (now.val === p.val) {
+      pRoute = [...stack, now];
+      continue;
+    }
+  }
+  const shortLength =
+    pRoute.length > qRoute.length ? qRoute.length : pRoute.length;
+  for (let i = 0; i < shortLength; i++) {
+    if (pRoute.at(i) !== qRoute.at(i)) {
+      answer = pRoute.at(i - 1);
+      break;
+    }
+  }
+  return answer ? answer : pRoute.at(shortLength - 1);
+}