Question

968. Binary Tree Cameras

中文文档

Description

You are given the root of a binary tree. We install cameras on the tree nodes where each camera at a node can monitor its parent, itself, and its immediate children.

Return _the minimum number of cameras needed to monitor all nodes of the tree_.

 

Example 1:

Input: root = [0,0,null,0,0]
Output: 1
Explanation: One camera is enough to monitor all nodes if placed as shown.

Example 2:

Input: root = [0,0,null,0,null,0,null,null,0]
Output: 2
Explanation: At least two cameras are needed to monitor all nodes of the tree. The above image shows one of the valid configurations of camera placement.

 

Constraints:

• The number of nodes in the tree is in the range [1, 1000].
• Node.val == 0

Solutions

Solution 1

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, val=0, left=None, right=None):
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def minCameraCover(self, root: Optional[TreeNode]) -> int:
    def dfs(root):
      if root is None:
        return inf, 0, 0
      la, lb, lc = dfs(root.left)
      ra, rb, rc = dfs(root.right)
      a = min(la, lb, lc) + min(ra, rb, rc) + 1
      b = min(la + rb, lb + ra, la + ra)
      c = lb + rb
      return a, b, c

    a, b, _ = dfs(root)
    return min(a, b)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode() {}
 *   TreeNode(int val) { this.val = val; }
 *   TreeNode(int val, TreeNode left, TreeNode right) {
 *     this.val = val;
 *     this.left = left;
 *     this.right = right;
 *   }
 * }
 */
class Solution {
  public int minCameraCover(TreeNode root) {
    int[] ans = dfs(root);
    return Math.min(ans[0], ans[1]);
  }

  private int[] dfs(TreeNode root) {
    if (root == null) {
      return new int[] {1 << 29, 0, 0};
    }
    var l = dfs(root.left);
    var r = dfs(root.right);
    int a = 1 + Math.min(Math.min(l[0], l[1]), l[2]) + Math.min(Math.min(r[0], r[1]), r[2]);
    int b = Math.min(Math.min(l[0] + r[1], l[1] + r[0]), l[0] + r[0]);
    int c = l[1] + r[1];
    return new int[] {a, b, c};
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *   TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *   TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
struct Status {
  int a, b, c;
};

class Solution {
public:
  int minCameraCover(TreeNode* root) {
    auto [a, b, _] = dfs(root);
    return min(a, b);
  }

  Status dfs(TreeNode* root) {
    if (!root) {
      return {1 << 29, 0, 0};
    }
    auto [la, lb, lc] = dfs(root->left);
    auto [ra, rb, rc] = dfs(root->right);
    int a = 1 + min({la, lb, lc}) + min({ra, rb, rc});
    int b = min({la + ra, la + rb, lb + ra});
    int c = lb + rb;
    return {a, b, c};
  };
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func minCameraCover(root *TreeNode) int {
  var dfs func(*TreeNode) (int, int, int)
  dfs = func(root *TreeNode) (int, int, int) {
    if root == nil {
      return 1 << 29, 0, 0
    }
    la, lb, lc := dfs(root.Left)
    ra, rb, rc := dfs(root.Right)
    a := 1 + min(la, min(lb, lc)) + min(ra, min(rb, rc))
    b := min(la+ra, min(la+rb, lb+ra))
    c := lb + rb
    return a, b, c
  }
  a, b, _ := dfs(root)
  return min(a, b)
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *   val: number
 *   left: TreeNode | null
 *   right: TreeNode | null
 *   constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *   }
 * }
 */

function minCameraCover(root: TreeNode | null): number {
  const dfs = (root: TreeNode | null): number[] => {
    if (!root) {
      return [1 << 29, 0, 0];
    }
    const [la, lb, lc] = dfs(root.left);
    const [ra, rb, rc] = dfs(root.right);
    const a = 1 + Math.min(la, lb, lc) + Math.min(ra, rb, rc);
    const b = Math.min(la + ra, la + rb, lb + ra);
    const c = lb + rb;
    return [a, b, c];
  };
  const [a, b, _] = dfs(root);
  return Math.min(a, b);
}