Update 261. Graph Valid Tree.md

Author: 0xkookoo

docs/Leetcode_Solutions/Python/261. Graph Valid Tree.md

@@ -1,62 +1,71 @@
-### 261. Graph Valid Tree
+# 261. Graph Valid Tree
 
+**<font color=red>难度: Medium</font>**
 
+## 刷题内容
 
-题目： 
-<https://leetcode.com/problems/graph-valid-tree/>
+> 原题连接
 
+* https://leetcode.com/problems/graph-valid-tree/
 
+> 内容描述
 
-难度 : Medium
+```
+Given n nodes labeled from 0 to n-1 and a list of undirected edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.
 
+Example 1:
 
+Input: n = 5, and edges = [[0,1], [0,2], [0,3], [1,4]]
+Output: true
+Example 2:
 
-思路：
+Input: n = 5, and edges = [[0,1], [1,2], [2,3], [1,3], [1,4]]
+Output: false
+Note: you can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0,1] is the same as [1,0] and thus will not appear together in edges.
+```
 
-graph 为 tree 两个条件：
+## 解题方案
 
-- 这个图是connected
-- 没有cycle
+> 思路 1
+******- 时间复杂度: O(V * E)******- 空间复杂度: O(V)******
 
+graph 为 tree 两个条件：
 
+- 这个图完全connected
+- 没有cycle
 
 偷懒AC代码，直接在323题，Number of Connected Components in an Undirected Graph上改的AC代码：
 
 
 
-```
-class Solution(object):
+```python
+class Solution:
     def validTree(self, n, edges):
         """
         :type n: int
         :type edges: List[List[int]]
         :rtype: bool
         """
         def find(x):
-        	if uf[x] != x:
-        		uf[x] = find(uf[x])
-        	return uf[x]
+            while x != uf[x]:
+                uf[x] = uf[uf[x]]
+                x = uf[x]
+            return uf[x]
 
-        def union(x,y):
-        	xRoot = find(x)
-        	yRoot = find(y)
-        	uf[xRoot] = yRoot
+        def union(x, y):
+            x_root = find(x)
+            y_root = find(y)
+            uf[x_root] = y_root
 
         uf = [i for i in range(n)]
 
         for node1, node2 in edges:
-            # cycle exists
-            if find(node1) == find(node2):
-                print 'ha '
+            if find(node1) == find(node2): # cycle exists
                 return False
             else:
                 union(node1, node2)
 
-        res = set()
-        for i in range(n):
-        	res.add(find(i))
-
-        return len(res) == 1
+        return len(set(find(i) for i in uf)) == 1 # fully connected
 ```