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| 1 | +package code; |
| 2 | +/* |
| 3 | + * 968. Binary Tree Cameras |
| 4 | + * 题意:在树上放置camera, 覆盖整棵树,一个camera可以监视自身,父节点和孩子节点,问最少camera数量 |
| 5 | + * 难度:Hard |
| 6 | + * 分类:Dynamic Programming, Tree, Depth-first Search |
| 7 | + * 思路:贪心算法,尽量让父节点放置 |
| 8 | + * Tips: |
| 9 | + */ |
| 10 | +public class lc968 { |
| 11 | + public class TreeNode { |
| 12 | + int val; |
| 13 | + TreeNode left; |
| 14 | + TreeNode right; |
| 15 | + TreeNode(int x) { val = x; } |
| 16 | + } |
| 17 | + int res; |
| 18 | + public int minCameraCover(TreeNode root) { |
| 19 | + if(root==null) return 0; |
| 20 | + int flag = dfs(root); |
| 21 | + if(flag==3) return res+1; //如果是3,代表root没被覆盖 |
| 22 | + return res; |
| 23 | + } |
| 24 | + |
| 25 | + public int dfs(TreeNode tn){ |
| 26 | + // 1代表被覆盖了,但该点没有camera。left,right都不为3的话,贪心算法,父节点不需要放置 |
| 27 | + // 2代表被覆盖了,该点有camera,父节点不需要放置 |
| 28 | + // 3代表没有被覆盖,所以父节点要放置 |
| 29 | + if(tn==null) return 1; |
| 30 | + int left = dfs(tn.left); |
| 31 | + int right = dfs(tn.right); |
| 32 | + if( left==3 || right==3){ |
| 33 | + res+=1; //在该节点放置 |
| 34 | + return 2; |
| 35 | + } |
| 36 | + if( left==2 || right==2 ) return 1; |
| 37 | + return 3; |
| 38 | + } |
| 39 | +} |
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