Google – Family Tree


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

家庭树判断亲属关系


第二轮:
输入一些 parent-child pairs, 然后给一个pair, 判断他们是否genetically related. 比如(p1, c1)表示p1生了c1, c1有p1的gene. 输入(p1, c1), (p2, c1), (p1, c2), (p2, c2). 画图可知(c1, c2)是related, (p1, p2)不是related. 面试官给了一个无向图, 结点就是p1, p2, c1, c2, 根据输入把边加上, 然后我就一直按无向图来做了, 觉得从一点dfs遍历能看到另一个点就行了. 其实这应该是有向图. 如果是无向图, p1通过c1可以到达p2, 但其实不应该. 有向图就不会有问题. 后来发现这其实很像树, 这题就是要找common ancestor, 如果有就是related. 把一个点的所有ancestor放到一个set里, 另一个点的放到另一个set里, 有交集就代表related. 这题最大的失误就是想复杂了, 想成general graph了, 其实根据遗传的知识来想要简单一些, 更像是树. 代码写出来了, 面试官说有bug, 但我觉得不是. 由于下一个面试官已经在外面等着了, 就没继续解释了. 事后想想应该说明白, 也就一两分钟的事儿, 面过试的应该都能理解. 最好还是不要有bug吧?

Family Tree
在family tree中找两个人是否是亲戚。follow up如果是亲戚的话,找lowest common ancestor。
第三题大概这样
class Node {
int val,
Node father;
Node mother;
}
public boolean isRelated(Node p1, Node p2);
follow up:
public Node findLowestCommonAncester(Node p1, Node p2);
其实这题有很多followup:
  1. 假设现在人分布于各个国家,比如,中国,澳大利亚,美国,那么有什么优化可以做
    1. 这是我当年的onsite问题,估计凉在这里
Example :
楼主做法 : 开始我考虑是parent指向child,然后稍微一比划不对,没让我多想面试官提示我反过来,让child指向parent,这样就是求两个node到leaf的路径中有没有交集了,dfs做出来。
帖子: https://www.1point3acres.com/bbs/thread-201712-1-1.html
路人建议 :以这个人为root节点,以父母为left和right,向上建树就是二叉树了
然后query的两个人就是两个root,对这两个二叉树进行BFS,找到最浅的公共节点n,n在tree1z和tree2的深度相加即可
这个follow up的一个关键点是找到lowest common ancestor就停了,所以用BFS比较好,不要把某一个人的ancestors都找出来了再去跟另一个人的ancestor找match.

Code:
Provider: null
class Node {
int val,
Node father;
Node mother;
}
public boolean isRelated(Node p1, Node p2) {
Set<Node> ancesters = new HashSet<>();
// BFS
Queue<Node> queue = new LinkedList<>();
queue.offer(p1);
while(!queue.isEmpty()) {
Node curr = queue.poll();
if(curr.father != null) queue.offer(curr.father);
if(curr.mother != null) queue.offer(curr.mother);
ancesters.add(curr);
}
queue = new LinkedList<>();
queue.offer(p2);
while(!queue.isEmpty()) {
Node curr = queue.poll();
if(ancesters.contains(curr)) return true;
if(curr.father != null) queue.offer(curr.father);
if(curr.mother != null) queue.offer(curr.mother);
}
return false;
}

follow up:
public Node findLowestCommonAncester(Node p1, Node p2) {
Set<Node> ancesters = new HashSet<>();
// BFS
Queue<Node> queue = new LinkedList<>();
queue.offer(p1);
while(!queue.isEmpty()) {
Node curr = queue.poll();
if(curr.father != null) queue.offer(curr.father);
if(curr.mother != null) queue.offer(curr.mother);
ancesters.add(curr);
}
queue = new LinkedList<>();
queue.offer(p2);
while(!queue.isEmpty()) {
Node curr = queue.poll();
if(ancesters.contains(curr)) return curr;
if(curr.father != null) queue.offer(curr.father);
if(curr.mother != null) queue.offer(curr.mother);
}
return false;

}

Google – Family Tree
family tree就是左右指针分别指向自己的父亲和母亲。最小一代就是root,leaves是最老的一代。
题目要求:
1. 给一个Family tree, 给每个节点标出对应binary heap的id
2. 任意给一个节点(没有root的情况下),输出这个点所在的层数
3. 任意给一个节点(没有root的情况下),输出这点和root的关系,比如是root的父亲的母亲就输出"FM"
[Solution]
三个问题其实就一个考点,就是看你知不知道binary tree用array的表示形式。
left child = 2 * id + 1
right child = 2 * id + 2
parent = (id – 1) / 2

class Solution {
   
  public void setID(TreeNode root) {
    dfs(root, 0);
  }
   
  private void dfs(TreeNode curr, int val) {
    if (curr == null) {
      return;
    }
    curr.val = val;
    dfs(curr.left, val * 2 + 1);
    dfs(curr.right, val * 2 + 2);
  }
   
  public int getLevel(TreeNode node) {
    int level = 0;
    while ((1 << level) < node.val) {
      level++;
    }
    return level;
  }
   
  public String getRelations(TreeNode node) {
    StringBuilder sb = new StringBuilder();
    getParent((node.val - 1) / 2, node.val, sb);
    return sb.reverse().toString();
  }
   
  private void getParent(int parentId, int id, StringBuilder sb) {
    if (id == 0) {
      return;
    }
    if (parentId * 2 + 1 == id) {
      sb.append("F");
    }
    else {
      sb.append("M");
    }
     
    getParent((parentId - 1) / 2, parentId, sb);
  }
}
Read full article from Google – Family Tree

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