Merge pull request #553 from Sunjae95/main

[선재] WEEK11 Solutions

Author: SamTheKorean

maximum-depth-of-binary-tree/sunjae95.js

@@ -0,0 +1,31 @@
+/**
+ * @description
+ * 최대 깊이를 탐색하는 문제여서 dfs를 먼저 떠올렸지만 이번에 bfs로 풀이하고 싶어 bfs로 풀었습니다.
+ *
+ * n = total node count
+ * time complexity: O(n)
+ * space complexity: O(n)
+ */
+var maxDepth = function (root) {
+  const queue = [];
+  let lastIndex = queue.length;
+  let answer = 0;
+
+  if (!root) return answer;
+
+  queue.push(root);
+
+  while (queue.length !== lastIndex) {
+    let currentCount = queue.length - lastIndex;
+    answer++;
+
+    while (currentCount--) {
+      let currentNode = queue[lastIndex];
+      lastIndex++;
+      if (currentNode.left) queue.push(currentNode.left);
+      if (currentNode.right) queue.push(currentNode.right);
+    }
+  }
+
+  return answer;
+};

reorder-list/sunjae95.js

@@ -0,0 +1,41 @@
+/**
+ * @description
+ * 인덱스로 접근하지 못하는 구조를 인덱스로 접근하게 하여 two pointer로 풀이
+ *
+ * n = total node count
+ * time complexity: O(n)
+ * space complexity: O(n)
+ */
+var reorderList = function (head) {
+  // convert from queue to list
+  let travelNode = head;
+  const list = [];
+  while (travelNode) {
+    list.push(travelNode);
+    travelNode = travelNode.next;
+  }
+  // two pointer
+  let [left, right] = [0, list.length - 1];
+  const node = new ListNode();
+  let tail = node;
+
+  while (left <= right) {
+    // 1. left append
+    const leftNode = list[left];
+    leftNode.next = null;
+    tail.next = leftNode;
+    tail = leftNode;
+    // 2. conditional right append
+    const rightNode = list[right];
+    rightNode.next = null;
+    if (left !== right) {
+      tail.next = rightNode;
+      tail = rightNode;
+    }
+
+    left++;
+    right--;
+  }
+
+  head = node.next;
+};