[LeetCode][JavaScript]Minimum Height Trees

Minimum Height Trees

For a undirected graph with tree characteristics, we can choose any node as the root. The result graph is then a rooted tree. Among all possible rooted trees, those with minimum height are called minimum height trees (MHTs). Given such a graph, write a function to find all the MHTs and return a list of their root labels.

Format
The graph contains n nodes which are labeled from 0 to n - 1. You will be given the number n and a list of undirected edges (each edge is a pair of labels).

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.

Example 1:

Given n = 4, edges = [[1, 0], [1, 2], [1, 3]]

        0
        |
        1
       /       2   3

return [1]

Example 2:

Given n = 6, edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]

     0  1  2
      \ | /
        3
        |
        4
        |
        5

return [3, 4]

https://leetcode.com/problems/minimum-height-trees/

可以转化为求从叶子到叶子最长的路径的中心，结果只可能是一个或者两个。

建立树，bfs遍历，每一轮都去掉所有的叶子节点，最后留下的就是结果。

开一个变量visited记录遍历过点的数量，如果visited >= n -2说明找到结果了。

 1 /**
 2  * @param {number} n
 3  * @param {number[][]} edges
 4  * @return {number[]}
 5  */
 6 var findMinHeightTrees = function(n, edges) {
 7     if(n === 1) return [0];
 8     var result = [], tree = {}, list = [], i, j, curr, visited = 0;
 9     for(i = 0; i < edges.length; i++){
10         curr = edges[i];
11         if(!tree[curr[0]]) tree[curr[0]] = new Node(curr[0]);
12         if(!tree[curr[1]]) tree[curr[1]] = new Node(curr[1]);
13         tree[curr[0]].neighbor.push(tree[curr[1]]);
14         tree[curr[1]].neighbor.push(tree[curr[0]]);
15     }
16     for(i in tree){
17         if(tree[i].neighbor.length === 1){
18             list.push(tree[i].val);
19         }
20     }
21     bfs(list);
22     for(i = 0; i < list.length; i++){
23         result.push(list[i]);
24     }
25     return result;
26
27     function Node(val){
28         this.val = val;
29         this.neighbor = [];
30     }
31     function bfs(list){
32         var len = list.length, top, topNeighbor;
33         if(visited >= n - 2) return;
34         while(len--){
35             visited++;
36             top = tree[list.shift()];
37             topNeighbor = top.neighbor[0];
38             deleteNode(topNeighbor.neighbor, top.val);
39             if(topNeighbor.neighbor.length <= 1 && list.indexOf(topNeighbor.val) === -1){
40                 list.push(topNeighbor.val);
41             }
42             delete tree[top.val];
43         }
44         bfs(list);
45     }
46     function deleteNode(arr, val){
47         for(var i = 0; i < arr.length; i++){
48             if(arr[i].val === val){
49                 arr.splice(i,1);
50                 return;
51             }
52         }
53     }
54 };