[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/

---

题意是以任意的节点为root，求出最高的那棵树。

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。

/**
 * @param {number} n
 * @param {number[][]} edges
 * @return {number[]}
 */
var findMinHeightTrees = function(n, edges) {
    if(n === 1) return [0];
    var result = [], tree = {}, list = [], i, j, curr, visited = 0;
    for(i = 0; i < edges.length; i++){
        curr = edges[i];
        if(!tree[curr[0]]) tree[curr[0]] = new Node(curr[0]);
        if(!tree[curr[1]]) tree[curr[1]] = new Node(curr[1]);
        tree[curr[0]].neighbor.push(tree[curr[1]]);
        tree[curr[1]].neighbor.push(tree[curr[0]]);
    }
    for(i in tree){
        if(tree[i].neighbor.length === 1){
            list.push(tree[i].val);
        }
    }
    bfs(list);
    for(i = 0; i < list.length; i++){
        result.push(list[i]);
    }
    return result;
    
    function Node(val){
        this.val = val;
        this.neighbor = [];
    }
    function bfs(list){
        var len = list.length, top, topNeighbor;
        if(visited >= n - 2) return;
        while(len--){
            visited++;
            top = tree[list.shift()];
            topNeighbor = top.neighbor[0];
            deleteNode(topNeighbor.neighbor, top.val);
            if(topNeighbor.neighbor.length <= 1 && list.indexOf(topNeighbor.val) === -1){
                list.push(topNeighbor.val);
            }
            delete tree[top.val];
        }
        bfs(list);
    }
    function deleteNode(arr, val){
        for(var i = 0; i < arr.length; i++){
            if(arr[i].val === val){
                arr.splice(i,1);
                return;
            }
        }
    }
};