Question

261. Graph Valid Tree

中文文档

Description

You have a graph of n nodes labeled from 0 to n - 1. You are given an integer n and a list of edges where edges[i] = [ai, bi] indicates that there is an undirected edge between nodes ai and bi in the graph.

Return true _if the edges of the given graph make up a valid tree, and_ false _otherwise_.

 

Example 1:

Input: n = 5, edges = [[0,1],[0,2],[0,3],[1,4]]
Output: true

Example 2:

Input: n = 5, edges = [[0,1],[1,2],[2,3],[1,3],[1,4]]
Output: false

 

Constraints:

• 1 <= n <= 2000
• 0 <= edges.length <= 5000
• edges[i].length == 2
• 0 <= ai, bi < n
• ai != bi
• There are no self-loops or repeated edges.

Solutions

Solution 1

class Solution:
  def validTree(self, n: int, edges: List[List[int]]) -> bool:
    def find(x):
      if p[x] != x:
        p[x] = find(p[x])
      return p[x]

    p = list(range(n))
    for a, b in edges:
      if find(a) == find(b):
        return False
      p[find(a)] = find(b)
      n -= 1
    return n == 1

class Solution {
  private int[] p;

  public boolean validTree(int n, int[][] edges) {
    p = new int[n];
    for (int i = 0; i < n; ++i) {
      p[i] = i;
    }
    for (int[] e : edges) {
      int a = e[0], b = e[1];
      if (find(a) == find(b)) {
        return false;
      }
      p[find(a)] = find(b);
      --n;
    }
    return n == 1;
  }

  private int find(int x) {
    if (p[x] != x) {
      p[x] = find(p[x]);
    }
    return p[x];
  }
}

class Solution {
public:
  vector<int> p;

  bool validTree(int n, vector<vector<int>>& edges) {
    p.resize(n);
    for (int i = 0; i < n; ++i) p[i] = i;
    for (auto& e : edges) {
      int a = e[0], b = e[1];
      if (find(a) == find(b)) return 0;
      p[find(a)] = find(b);
      --n;
    }
    return n == 1;
  }

  int find(int x) {
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
  }
};

func validTree(n int, edges [][]int) bool {
  p := make([]int, n)
  for i := range p {
    p[i] = i
  }
  var find func(x int) int
  find = func(x int) int {
    if p[x] != x {
      p[x] = find(p[x])
    }
    return p[x]
  }
  for _, e := range edges {
    a, b := e[0], e[1]
    if find(a) == find(b) {
      return false
    }
    p[find(a)] = find(b)
    n--
  }
  return n == 1
}

/**
 * @param {number} n
 * @param {number[][]} edges
 * @return {boolean}
 */
var validTree = function (n, edges) {
  let p = new Array(n);
  for (let i = 0; i < n; ++i) {
    p[i] = i;
  }
  function find(x) {
    if (p[x] != x) {
      p[x] = find(p[x]);
    }
    return p[x];
  }
  for (const [a, b] of edges) {
    if (find(a) == find(b)) {
      return false;
    }
    p[find(a)] = find(b);
    --n;
  }
  return n == 1;
};