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Commit 3de8be6

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feat: add solutions to lcp problem: No.64
1 parent 9fcd034 commit 3de8be6

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‎lcp/LCP 64. 二叉树灯饰/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:递归**
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我们注意到,三个开关只能影响当前节点及其左右子节点,因此我们可以将当前节点的状态分为四种:
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- 全灭:当前节点及其左右子节点的灯均处于关闭状态;
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- 全亮:当前节点及其左右子节点的灯均处于开启状态;
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- 当前灯亮:当前节点的灯处于开启状态,其余节点的灯均处于关闭状态;
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- 当前灯灭:当前节点的灯处于关闭状态,其余节点的灯均处于开启状态;
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我们用 $t_1,ドル $t_2,ドル $t_3,ドル $t_4$ 分别表示四种状态下,需要操作的开关次数。我们可以发现,对于当前节点的状态,我们先递归计算其左右子节点的状态 $l_1,ドル $l_2,ドル $l_3,ドル $l_4$ 和 $r_1,ドル $r_2,ドル $r_3,ドル $r_4,ドル然后根据当前节点的灯的状态,可以得到四种状态下的最小操作次数:
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如果当前节点的灯处于开启状态,那么:
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- 全灭 $t_1 = min(l_1 + r_1 + 1, l_2 + r_2 + 1, l_3 + r_3 + 1, l_4 + r_4 + 3)$
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- 全亮 $t_2 = min(l_1 + r_1 + 2, l_2 + r_2, l_3 + r_3 + 2, l_4 + r_4 + 2)$
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- 当前灯亮 $t_3 = min(l_1 + r_1, l_2 + r_2 + 2, l_3 + r_3 + 2, l_4 + r_4 + 2)$
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- 当前灯灭 $t_4 = min(l_1 + r_1 + 1, l_2 + r_2 + 1, l_3 + r_3 + 3, l_4 + r_4 + 1)$
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如果当前节点的灯处于关闭状态,那么:
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- 全灭 $t_1 = min(l_1 + r_1, l_2 + r_2 + 2, l_3 + r_3 + 2, l_4 + r_4 + 2)$
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- 全亮 $t_2 = min(l_1 + r_1 + 1, l_2 + r_2 + 1, l_3 + r_3 + 3, l_4 + r_4 + 1)$
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- 当前灯亮 $t_3 = min(l_1 + r_1 + 1, l_2 + r_2 + 1, l_3 + r_3 + 1, l_4 + r_4 + 3)$
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- 当前灯灭 $t_4 = min(l_1 + r_1 + 2, l_2 + r_2, l_3 + r_3 + 2, l_4 + r_4 + 2)$
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最后,我们返回四种状态下的最小操作次数即可。
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最终答案为 $t_1,ドル因为我们需要将所有节点的灯都关闭。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 为二叉树的节点个数。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution:
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def closeLampInTree(self, root: TreeNode) -> int:
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def dfs(root):
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if root is None:
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return 0, 0, 0, 0
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l1, l2, l3, l4 = dfs(root.left)
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r1, r2, r3, r4 = dfs(root.right)
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t1 = t2 = t3 = t4 = inf
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if root.val:
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t1 = min(l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 1, l4 + r4 + 3)
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t2 = min(l1 + r1 + 2, l2 + r2, l3 + r3 + 2, l4 + r4 + 2)
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t3 = min(l1 + r1, l2 + r2 + 2, l3 + r3 + 2, l4 + r4 + 2)
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t4 = min(l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 3, l4 + r4 + 1)
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else:
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t1 = min(l1 + r1, l2 + r2 + 2, l3 + r3 + 2, l4 + r4 + 2)
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t2 = min(l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 3, l4 + r4 + 1)
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t3 = min(l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 1, l4 + r4 + 3)
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t4 = min(l1 + r1 + 2, l2 + r2, l3 + r3 + 2, l4 + r4 + 2)
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return t1, t2, t3, t4
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return dfs(root)[0]
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public int closeLampInTree(TreeNode root) {
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return dfs(root)[0];
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}
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private int[] dfs(TreeNode root) {
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int[] ans = new int[4];
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if (root == null) {
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return ans;
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}
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int[] left = dfs(root.left);
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int[] right = dfs(root.right);
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int l1 = left[0], l2 = left[1], l3 = left[2], l4 = left[3];
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int r1 = right[0], r2 = right[1], r3 = right[2], r4 = right[3];
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if (root.val != 0) {
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ans[0] = min(l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 1, l4 + r4 + 3);
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ans[1] = min(l1 + r1 + 2, l2 + r2, l3 + r3 + 2, l4 + r4 + 2);
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ans[2] = min(l1 + r1, l2 + r2 + 2, l3 + r3 + 2, l4 + r4 + 2);
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ans[3] = min(l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 3, l4 + r4 + 1);
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} else {
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ans[0] = min(l1 + r1, l2 + r2 + 2, l3 + r3 + 2, l4 + r4 + 2);
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ans[1] = min(l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 3, l4 + r4 + 1);
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ans[2] = min(l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 1, l4 + r4 + 3);
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ans[3] = min(l1 + r1 + 2, l2 + r2, l3 + r3 + 2, l4 + r4 + 2);
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}
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return ans;
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}
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private int min(int... nums) {
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int ans = 1 << 30;
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for (int num : nums) {
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ans = Math.min(ans, num);
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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int closeLampInTree(TreeNode* root) {
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return dfs(root)[0];
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}
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vector<int> dfs(TreeNode* root) {
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vector<int> ans(4);
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if (!root) {
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return ans;
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}
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auto left = dfs(root->left);
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auto right = dfs(root->right);
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int l1 = left[0], l2 = left[1], l3 = left[2], l4 = left[3];
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int r1 = right[0], r2 = right[1], r3 = right[2], r4 = right[3];
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if (root->val) {
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ans[0] = min({l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 1, l4 + r4 + 3});
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ans[1] = min({l1 + r1 + 2, l2 + r2, l3 + r3 + 2, l4 + r4 + 2});
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ans[2] = min({l1 + r1, l2 + r2 + 2, l3 + r3 + 2, l4 + r4 + 2});
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ans[3] = min({l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 3, l4 + r4 + 1});
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} else {
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ans[0] = min({l1 + r1, l2 + r2 + 2, l3 + r3 + 2, l4 + r4 + 2});
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ans[1] = min({l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 3, l4 + r4 + 1});
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ans[2] = min({l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 1, l4 + r4 + 3});
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ans[3] = min({l1 + r1 + 2, l2 + r2, l3 + r3 + 2, l4 + r4 + 2});
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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/**
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* Definition for a binary tree node.
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* type TreeNode struct {
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* Val int
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* Left *TreeNode
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* Right *TreeNode
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* }
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*/
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func closeLampInTree(root *TreeNode) (ans int) {
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const inf = 1 << 30
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var dfs func(*TreeNode) (int, int, int, int)
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dfs = func(root *TreeNode) (int, int, int, int) {
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if root == nil {
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return 0, 0, 0, 0
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}
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l1, l2, l3, l4 := dfs(root.Left)
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r1, r2, r3, r4 := dfs(root.Right)
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t1, t2, t3, t4 := inf, inf, inf, inf
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if root.Val == 1 {
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t1 = min(l1+r1+1, l2+r2+1, l3+r3+1, l4+r4+3)
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t2 = min(l1+r1+2, l2+r2, l3+r3+2, l4+r4+2)
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t3 = min(l1+r1, l2+r2+2, l3+r3+2, l4+r4+2)
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t4 = min(l1+r1+1, l2+r2+1, l3+r3+3, l4+r4+1)
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} else {
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t1 = min(l1+r1, l2+r2+2, l3+r3+2, l4+r4+2)
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t2 = min(l1+r1+1, l2+r2+1, l3+r3+3, l4+r4+1)
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t3 = min(l1+r1+1, l2+r2+1, l3+r3+1, l4+r4+3)
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t4 = min(l1+r1+2, l2+r2, l3+r3+2, l4+r4+2)
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}
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return t1, t2, t3, t4
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}
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ans, _, _, _ = dfs(root)
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return
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}
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func min(a, b, c, d int) int {
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if b < a {
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a = b
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}
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if c < a {
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a = c
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}
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if d < a {
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a = d
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}
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return a
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}
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```
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### **...**
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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int closeLampInTree(TreeNode* root) {
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return dfs(root)[0];
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}
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vector<int> dfs(TreeNode* root) {
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vector<int> ans(4);
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if (!root) {
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return ans;
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}
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auto left = dfs(root->left);
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auto right = dfs(root->right);
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int l1 = left[0], l2 = left[1], l3 = left[2], l4 = left[3];
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int r1 = right[0], r2 = right[1], r3 = right[2], r4 = right[3];
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if (root->val) {
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ans[0] = min({l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 1, l4 + r4 + 3});
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ans[1] = min({l1 + r1 + 2, l2 + r2, l3 + r3 + 2, l4 + r4 + 2});
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ans[2] = min({l1 + r1, l2 + r2 + 2, l3 + r3 + 2, l4 + r4 + 2});
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ans[3] = min({l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 3, l4 + r4 + 1});
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} else {
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ans[0] = min({l1 + r1, l2 + r2 + 2, l3 + r3 + 2, l4 + r4 + 2});
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ans[1] = min({l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 3, l4 + r4 + 1});
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ans[2] = min({l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 1, l4 + r4 + 3});
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ans[3] = min({l1 + r1 + 2, l2 + r2, l3 + r3 + 2, l4 + r4 + 2});
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}
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return ans;
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}
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};

‎lcp/LCP 64. 二叉树灯饰/Solution.go

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/**
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* Definition for a binary tree node.
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* type TreeNode struct {
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* Val int
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* Left *TreeNode
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* Right *TreeNode
7+
* }
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*/
9+
func closeLampInTree(root *TreeNode) (ans int) {
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const inf = 1 << 30
11+
var dfs func(*TreeNode) (int, int, int, int)
12+
dfs = func(root *TreeNode) (int, int, int, int) {
13+
if root == nil {
14+
return 0, 0, 0, 0
15+
}
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l1, l2, l3, l4 := dfs(root.Left)
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r1, r2, r3, r4 := dfs(root.Right)
18+
t1, t2, t3, t4 := inf, inf, inf, inf
19+
if root.Val == 1 {
20+
t1 = min(l1+r1+1, l2+r2+1, l3+r3+1, l4+r4+3)
21+
t2 = min(l1+r1+2, l2+r2, l3+r3+2, l4+r4+2)
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t3 = min(l1+r1, l2+r2+2, l3+r3+2, l4+r4+2)
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t4 = min(l1+r1+1, l2+r2+1, l3+r3+3, l4+r4+1)
24+
} else {
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t1 = min(l1+r1, l2+r2+2, l3+r3+2, l4+r4+2)
26+
t2 = min(l1+r1+1, l2+r2+1, l3+r3+3, l4+r4+1)
27+
t3 = min(l1+r1+1, l2+r2+1, l3+r3+1, l4+r4+3)
28+
t4 = min(l1+r1+2, l2+r2, l3+r3+2, l4+r4+2)
29+
}
30+
return t1, t2, t3, t4
31+
}
32+
ans, _, _, _ = dfs(root)
33+
return
34+
}
35+
36+
func min(a, b, c, d int) int {
37+
if b < a {
38+
a = b
39+
}
40+
if c < a {
41+
a = c
42+
}
43+
if d < a {
44+
a = d
45+
}
46+
return a
47+
}
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Original file line numberDiff line numberDiff line change
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
5+
* TreeNode left;
6+
* TreeNode right;
7+
* TreeNode(int x) { val = x; }
8+
* }
9+
*/
10+
class Solution {
11+
public int closeLampInTree(TreeNode root) {
12+
return dfs(root)[0];
13+
}
14+
15+
private int[] dfs(TreeNode root) {
16+
int[] ans = new int[4];
17+
if (root == null) {
18+
return ans;
19+
}
20+
int[] left = dfs(root.left);
21+
int[] right = dfs(root.right);
22+
int l1 = left[0], l2 = left[1], l3 = left[2], l4 = left[3];
23+
int r1 = right[0], r2 = right[1], r3 = right[2], r4 = right[3];
24+
if (root.val != 0) {
25+
ans[0] = min(l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 1, l4 + r4 + 3);
26+
ans[1] = min(l1 + r1 + 2, l2 + r2, l3 + r3 + 2, l4 + r4 + 2);
27+
ans[2] = min(l1 + r1, l2 + r2 + 2, l3 + r3 + 2, l4 + r4 + 2);
28+
ans[3] = min(l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 3, l4 + r4 + 1);
29+
} else {
30+
ans[0] = min(l1 + r1, l2 + r2 + 2, l3 + r3 + 2, l4 + r4 + 2);
31+
ans[1] = min(l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 3, l4 + r4 + 1);
32+
ans[2] = min(l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 1, l4 + r4 + 3);
33+
ans[3] = min(l1 + r1 + 2, l2 + r2, l3 + r3 + 2, l4 + r4 + 2);
34+
}
35+
return ans;
36+
}
37+
38+
private int min(int... nums) {
39+
int ans = 1 << 30;
40+
for (int num : nums) {
41+
ans = Math.min(ans, num);
42+
}
43+
return ans;
44+
}
45+
}

‎lcp/LCP 64. 二叉树灯饰/Solution.py

Lines changed: 27 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,27 @@
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# Definition for a binary tree node.
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# class TreeNode:
3+
# def __init__(self, x):
4+
# self.val = x
5+
# self.left = None
6+
# self.right = None
7+
class Solution:
8+
def closeLampInTree(self, root: TreeNode) -> int:
9+
def dfs(root):
10+
if root is None:
11+
return 0, 0, 0, 0
12+
l1, l2, l3, l4 = dfs(root.left)
13+
r1, r2, r3, r4 = dfs(root.right)
14+
t1 = t2 = t3 = t4 = inf
15+
if root.val:
16+
t1 = min(l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 1, l4 + r4 + 3)
17+
t2 = min(l1 + r1 + 2, l2 + r2, l3 + r3 + 2, l4 + r4 + 2)
18+
t3 = min(l1 + r1, l2 + r2 + 2, l3 + r3 + 2, l4 + r4 + 2)
19+
t4 = min(l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 3, l4 + r4 + 1)
20+
else:
21+
t1 = min(l1 + r1, l2 + r2 + 2, l3 + r3 + 2, l4 + r4 + 2)
22+
t2 = min(l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 3, l4 + r4 + 1)
23+
t3 = min(l1 + r1 + 1, l2 + r2 + 1, l3 + r3 + 1, l4 + r4 + 3)
24+
t4 = min(l1 + r1 + 2, l2 + r2, l3 + r3 + 2, l4 + r4 + 2)
25+
return t1, t2, t3, t4
26+
27+
return dfs(root)[0]

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