Graph Valid Tree

Description

null

Solution(javascript)

// 1. 什么样的图是树？没有环而且不是森林
// 2. 无向图检测是否有环？DFS/Union Find


// /** 1: DFS
//  * @param {number} n
//  * @param {number[][]} edges
//  * @return {boolean}
//  */
// const validTree = function (n, edges) {
//   const adj = edges.reduce((acc, [a, b]) => {
//     acc[a] = acc[a] || []
//     acc[a].push(b)
//     acc[b] = acc[b] || []
//     acc[b].push(a)
//     return acc
//   }, {})
//   const root = 0
//   const parent = {
//     [root]: -1,
//   }

//   let size = 1
//   let hasCycle = false
//   const dfs = (node) => {
//     (adj[node] || []).forEach((next) => {
//       if (parent[next] === undefined) {
//         size += 1
//         parent[next] = node
//         dfs(next)
//       } else { // visited
//         if (next !== parent[node]) {
//           hasCycle = true
//         }
//       }
//     })
//   }
//   dfs(root)
//   return !hasCycle && size === n
// }


/** 1: Union Find
 * @param {number} n
 * @param {number[][]} edges
 * @return {boolean}
 */
const validTree = function (n, edges) {
  const parent = new Array(n).fill(0).map((x, index) => index)
  const size = new Array(n).fill(0).map(() => 1)
  let count = n

  const root = (a) => {
    while (a !== parent[a]) {
      parent[a] = parent[parent[a]]
      a = parent[a]
    }
    return a
  }

  const union = (a, b) => {
    const rootA = root(a)
    const rootB = root(b)
    if (rootA === rootB) {
      return
    }
    count -= 1
    if (size[rootA] < size[rootB]) {
      parent[rootA] = rootB
      size[rootB] += size[rootA]
    } else {
      parent[rootB] = rootA
      size[rootA] += size[rootB]
    }
  }
  const connected = (a, b) => root(a) === root(b)
  for (const [a, b] of edges) {
    if (connected(a, b)) {
      return false
    }
    union(a, b)
  }
  return count === 1
}