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문제 Given an m x n matrix board containing 'X' and 'O', capture all regions that are 4-directionally surrounded by 'X'. A region is captured by flipping all 'O's into 'X's in that surrounded region. Example 1: Input: board = [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]] Output: [["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]] Explanation: Notic..

문제 Given the root of a binary tree, return the inorder traversal of its nodes' values. Example 1: Input: root = [1,null,2,3] Output: [1,3,2] Example 2: Input: root = [] Output: [] Example 3: Input: root = [1] Output: [1] Constraints: The number of nodes in the tree is in the range [0, 100]. -100

문제 Given the root of a binary tree, return its maximum depth. A binary tree's maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. Example 1: Input: root = [3,9,20,null,null,15,7] Output: 3 Example 2: Input: root = [1,null,2] Output: 2 Constraints: The number of nodes in the tree is in the range [0, 104]. -100

class Node { constructor(data) { this.data = data; this.children = []; } add(data) { this.children.push(new Node(data)); } remove(data) { this.children = this.children.filter((child) => child.data === data ? false : true ); } } class Tree { constructor() { this.root = null; } BFS(fn) { if (this.root === null) return; const unvisitedQueue = [this.root]; while (unvisitedQueue.length !== 0) { const..