算法--监控二叉树

Posted 2021-04-09 前端学习记录簿

分析：对每个节点来说，为达到最小使用量，父节点可以监控所有的节点来，当子节点下没有更多的节点时，父节点安装监控即可，当子节点下有更多节点时，子节点安装即可。

思考如下：

function minCamera(tree){ let count = 0; let arr = [tree]; while(arr.length){ let node = arr.pop(); let left = node.left,right=node.right;    // 可一个子节点下还有节点时，可不关注另外一个节点下是否还有节点    if(left&&(left.left||left.right)&&right&&(right.left||right.right)){      count+=2;    }else if((left&&(left.left||left.right)&&right)||(right&&(right.left||right.right)&&left)){      count+=2;    }else if(left||right){     count+=1;    }else if(count==0){     count+=1;    }    if(left&&left.left){       arr.push(left.left);     }     if(left&&left.right){       arr.push(left.right);     }     if(right&&right.left){       arr.push(right.left);     }     if(right&&right.right){       arr.push(right.right);     } } return count;}

var ex1 = { left: { left: {}, right: {} }, right: null}var ex2 = { left: { left: { left: { right:{} } } }}var ex3={ left: { left: {} }, right: { left: {}, right: {} }}var ex4 = { left: { left: {} }, right: { left: {}, right: { right: {} } }}var ex5 = {
}var ex6 = { left: {}, right: {}}
var res1 = minCamera(ex1)var res2 = minCamera(ex2)var res3 = minCamera(ex3)var res4 = minCamera(ex4)var res5 = minCamera(ex5)var res6 = minCamera(ex6)
console.log(res1,res2,res3,res4,res5,res6);// 输出如下：1 2 2 3 1 1

其他参考解法：1、参考链接：leetcode 每日一题 968 监控二叉树_Mosiclone的博客-CSDN博客

class getCamera{  // 递归遍历整个树  // status有三个状态：0不能被监测到，1自身装了摄像头，2没装摄像头可以被监视  count = 0;  dfs(node){    if(!node) return 2; //null节点返回2，这样叶子节点的值j就是0    const left = this.dfs(node.left);    const right = this.dfs(node.right);    if(left ==0 ||right ==0){      this.count++;      return 1;    }    if(left ==1 || right==1) return 2;    return 0;  }  minCameraCover(root){   if(!root) return 0;    if(!root.right&&!root.left) return 1;    if(this.dfs(root)==0) this.count++;    return this.count;  }}