Generate Random Tree - Google


https://docs.google.com/document/d/1qxA2wps0IhVRWULulQ55W4SGPMu2AE5MkBB37h8Dr58/

构造一棵size为n的随机树


第三轮,三哥,题目就一句话,让我先随机构造一颗size为N的树。我开始没听到“random”一词,就以为构成一颗树就行。秉着论坛经验贴说的,要提前考虑各种条件。我就开始问他,有木有其他限制,比如形状有木有要求,普通的树还是二叉树,N<=0怎么办等等。。。他就说没有,只要size是N就行了

  0                   1
  /                   / \
 1                  2 0
/                      
2                    
如果要求上图是不同的两棵树(虽然它们选择的边都相同,但是根不同),可以用以下思路
  1. pool初始化为[1,2,3, … n]
  2. 从pool中随机选择一个vertex V准备加入tree
  3. 从tree中随机选择一个vertex U作为V的父亲节点
  4. 将v从pool中删除,加入u的孩子中
  5. 重复从2开始,直到pool为空
构造
  0  => P = ⅓
  /                   
 1    => P = ½               
/                      
2      => P = ½  
概率:P = 1/12
  0        0 0          0 root == 0,4 situations, total == 12 situations
 /          / /  \ / \
1          2 1  2 2 1
/          /
2         1
            
参考代码:
Provider: null
private class TreeNode{
int val;
List<TreeNode> children;
public TreeNode(int x) {
val = x;
children = new ArrayList<>();
}
}

public TreeNode getTree(int N) {
if(N == 0) return null;
List<TreeNode> pool = new ArrayList<>();
for(int i = 0 ; i < N; i++) {
pool.add(new TreeNode(i));
}
List<TreeNode> tree = new ArrayList<>();
Random rand = new Random();
while (pool.size() > 0) {
int idx = rand.nextInt(pool.size());
TreeNode curr = pool.get(idx);
if(tree.size() == 0) {
tree.add(curr);
pool.remove(idx);
} else {
int parent = rand.nextInt(tree.size());
tree.get(parent).children.add(curr);
tree.add(curr);
pool.remove(idx);
}
}
return tree.get(0);
}
Time complexity: o(n^2)


如果题目要求是构造一棵无根的label tree
  0                   1
  /                   / \
 1                  2 0
/                      
2      
一张n个顶点的联通图中构造一棵随机的生成树,上图两树被看作一棵树

思路: 先构造随机的Prüfer sequence,然后根据sequence生成树
(cayley’s formula中用到的树的表示方法)
Time complexity: O(n ^ 2)
参考代码
Provider: null
public Map<Integer, List<Integer>> getRandomTree(int n) {
    if(n <= 2) {
    Map<Integer, List<Integer>> tree = new HashMap<>();
    for(int i = 0 ; i < n ; i++) {
    tree.put(i, new ArrayList<>());
    }
    for(int i = 0 ; i < n ; i++) {
    for(int j = 0 ; j < n ; j++) {
    if(i != j) {
    tree.get(i).add(j);
    tree.get(j).add(i);
    }
    }
    }
    return tree;
    }
    int[] prufer = new int[n - 2];
    Random rand = new Random();
    for(int i = 0 ; i < prufer.length ; i++) {
    prufer[i] = rand.nextInt(n);
    }
    return getTree(prufer);
}

// 根据prufer sequence得到label tree
public Map<Integer, List<Integer>> getTree(int[] prufer) {
    int n = prufer.length + 2;
    Map<Integer, List<Integer>> tree = new HashMap<>();
    for(int i = 0 ; i < n ; i++) {
    tree.put(i, new ArrayList<>());
    }
    int[] vertices = new int[n];
    for(int x : prufer) {
    vertices[x]++;
    }
    for(int i = 0 ; i < prufer.length ; i++) {
    for(int j = 0 ; j < vertices.length ; j++) {
    if(vertices[j] == 0) {
    // connect j <--> prufer[i]
    tree.get(j).add(prufer[i]);
    tree.get(prufer[i]).add(j);
    vertices[j] = -1; // delete from vertex set
    vertices[prufer[i]]--; //delete from prufer set
    break;
    }
    }
    }
// connect the last two vertex with vertices[i] == 0
    Integer first = null;
    for(int i = 0 ; i < vertices.length ; i++) {
    if(vertices[i] == 0) {
    if(first == null) first = i;
    else {
    tree.get(first).add(i);
    tree.get(i).add(first);
    break;
    }
    }
    }
    return tree;
}

https://en.wikipedia.org/wiki/Pr%C3%BCfer_sequence

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