feat(soobing): week10 > invert-binary-tree

Author: soobing

invert-binary-tree/soobing.ts

@@ -0,0 +1,64 @@
+/**
+ * 문제 설명
+ * - 이진 트리를 반전시키는 문제
+ *
+ * 아이디어
+ * 1) DFS / BFS 로 탐색하면서 반전시키기
+ * - 시간 복잡도 O(n): 모든 노드 한번씩 방문
+ * - 공간 복잡도 DFS의 경우 O(h), BFS의 경우 O(2/n) -> 마지막 레벨의 노드 수
+ */
+
+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 invertTree(root: TreeNode | null): TreeNode | null {
+  if (!root) return null;
+
+  const left = invertTree(root.left);
+  const right = invertTree(root.right);
+
+  root.left = right;
+  root.right = left;
+
+  return root;
+}
+
+/**
+ * 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 invertTreeBFS(root: TreeNode | null): TreeNode | null {
+  const queue: (TreeNode | null)[] = [root];
+
+  while (queue.length > 0) {
+    const current = queue.shift();
+    if (current) {
+      const left = current.left;
+      const right = current.right;
+      current.left = right;
+      current.right = left;
+
+      queue.push(left);
+      queue.push(right);
+    }
+  }
+  return root;
+}