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:
![](https://spcdn.pages.dev/leetcode/problems/0261.Graph%20Valid%20Tree/images/tree1-graph.jpg)
Input: n = 5, edges = [[0,1],[0,2],[0,3],[1,4]]
Output: true
Example 2:
![](https://spcdn.pages.dev/leetcode/problems/0261.Graph%20Valid%20Tree/images/tree2-graph.jpg)
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#
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| 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
|
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| 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];
}
}
|
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| 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];
}
};
|
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| 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
}
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| /**
* @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;
};
|