binary tree level order traversal solution

Author: hyer0705

binary-tree-level-order-traversal/hyer0705.ts

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+/**
+ * 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)
+ *     }
+ * }
+ */
+
+// ----- [Solution 1]
+// Time Complexity: O(n)
+// Space Complexity: O(n)
+function levelOrder(root: TreeNode | null): number[][] {
+  if (!root) return [];
+
+  const answer: number[][] = [];
+
+  const queue: [TreeNode, number][] = [];
+  let pointer = 0;
+
+  queue.push([root, 0]);
+
+  while (pointer < queue.length) {
+    const [currNode, level] = queue[pointer++];
+
+    if (!answer[level]) answer[level] = [];
+
+    answer[level].push(currNode.val);
+
+    if (currNode.left) queue.push([currNode.left, level + 1]);
+    if (currNode.right) queue.push([currNode.right, level + 1]);
+  }
+
+  return answer;
+}
+
+// ----- [Solution 2]
+// Time Complexity: O(n)
+// Space Complexity: O(n)
+function levelOrder(root: TreeNode | null): number[][] {
+  if (!root) return [];
+
+  const answer: number[][] = [];
+
+  const queue: TreeNode[] = [];
+  queue.push(root);
+
+  let pointer = 0;
+
+  while (pointer < queue.length) {
+    const levelSize = queue.length - pointer;
+    const currentLevel: number[] = [];
+
+    for (let i = 0; i < levelSize; i++) {
+      const currNode = queue[pointer++];
+      currentLevel.push(currNode.val);
+
+      if (currNode.left) queue.push(currNode.left);
+      if (currNode.right) queue.push(currNode.right);
+    }
+
+    answer.push(currentLevel);
+  }
+
+  return answer;
+}