Easy to understand code with detailed explanation
The problem asks us to reorient some roads such that each city can visit the capital city (city 0) and return the minimum number of roads changed. We are given 
The problem asks us to reorient some roads such that each city can visit the capital city (city 0) and return the minimum number of roads changed. We are given n cities numbered from 0 to n-1 and n-1 roads that form a tree structure.\n\nTo solve this problem, we need to identify which roads need to be reversed to reach the capital city from every other city. We can start from the capital city (city 0) and traverse the tree structure to identify the required reversals. We can use a Depth-First Search (DFS) algorithm for the traversals.\n\nFirst, we initialize an object called `neighbors` to store the neighboring cities for each city, and a boolean array called `isZeroReachable` to track the cities that can reach the capital city.\n\nThen, we iterate over the connections array and populate the neighbors object with the neighboring cities for each city. For each connection, we add both the source city and the destination city to each other\'s neighboring list, with an indication of whether the direction of the connection needs to be reversed or not.\n\nNext, we start the DFS from the capital city (city 0) and then recursively traverse its neighbors and mark them as reachable, while counting the number of connections that need to be reversed.\n\nFinally, we return the total number of connections that need to be reversed.\n\n# JavaScript Code\n```\nvar minReorder = function (n, connections) {\n  // Create an object to store the neighbors of each city\n  let neighbors = {};\n  // Create an array to keep track of which cities can reach city 0\n  const isZeroReachable = Array(n).fill(false);\n  // City 0 can reach itself\n  isZeroReachable[0] = true;\n\n  // Initialize an empty array for each city in neighbors\n  for (let city = 0; city < n; city++) neighbors[city] = [];\n\n  for (const [sourceCity, destinationCity] of connections) {\n    // Add the destination city to the array for the source city with the boolean value false\n    // boolean tracks the direction of the connection\n    neighbors[sourceCity].push([destinationCity, false]);\n    // Add the source city to the array for the destination city with the boolean value true\n    neighbors[destinationCity].push([sourceCity, true]);\n  }\n\n  // Initialize the reorder count to 0 and call traverseCities with city 0\n  let reorderCount = 0;\n  traverseCities(0);\n\n  // Return the reorder count\n  return reorderCount;\n\n  /**\n   * @param {number} zeroReachableCity\n   */\n  function traverseCities(zeroReachableCity) {\n    // Iterate over each neighbor of zeroReachableCity\n    for (const [neighborCity, isSourceCity] of neighbors[zeroReachableCity]) {\n      // If the neighbor is already reachable, continue to the next neighbor\n      if (isZeroReachable[neighborCity]) continue;\n      // If the connection is outgoing, increment reorderCount\n      if (!isSourceCity) reorderCount++;\n      // Mark the neighbor as reachable\n      isZeroReachable[neighborCity] = true;\n      // Recursively call traverseCities with the neighbor city as the new zeroReachableCity\n      traverseCities(neighborCity);\n    }\n  }\n};\n\n```
['JavaScript']