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feat: update solutions to lc problems: No.2331,2974,3011 #3258

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Jul 11, 2024
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feat: update solutions to lc problems: No.2331,2974,3011 #3258

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202 changes: 41 additions & 161 deletions solution/2300-2399/2331.Evaluate Boolean Binary Tree/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -82,14 +82,12 @@ OR 运算节点的值为 True OR False = True 。

我们可以使用递归的方式来求解本题。

对于当前节点 `root`：
对于当前节点 $\textit{root}$：

- 如果其左右孩子都为空，说明是叶子节点，此时判断其值是否为 $1$，如果是，则返回 `true`，否则返回 `false`。
- 否则，对其左右孩子分别递归求解，得到其左右孩子的值 $l$ 和 $r$。然后根据当前节点值的不同，分别进行如下操作：
- 如果当前节点值为 $2$，则返回 `l or r`。
- 如果当前节点值为 $3$，则返回 `l && r`。
- 如果其左孩子为空，说明当前节点是叶子节点。如果当前节点的值为 $1$，则返回 $\text{true}$，否则返回 $\text{false}$；
- 如果当前节点的值为 $2$，则返回其左孩子和右孩子的递归结果的逻辑或，否则返回其左孩子和右孩子的递归结果的逻辑与。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树节点个数。
时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树的节点个数。

<!-- tabs:start -->

Expand All @@ -104,13 +102,10 @@ OR 运算节点的值为 True OR False = True 。
# self.right = right
class Solution:
def evaluateTree(self, root: Optional[TreeNode]) -> bool:
def dfs(root):
if root.left is None and root.right is None:
return bool(root.val)
l, r = dfs(root.left), dfs(root.right)
return (l or r) if root.val == 2 else (l and r)

return dfs(root)
if root.left is None:
return bool(root.val)
op = or_ if root.val == 2 else and_
return op(self.evaluateTree(root.left), self.evaluateTree(root.right))
```

#### Java
Expand All @@ -133,18 +128,13 @@ class Solution:
*/
class Solution {
public boolean evaluateTree(TreeNode root) {
return dfs(root);
}

private boolean dfs(TreeNode root) {
if (root.left == null && root.right == null) {
if (root.left == null) {
return root.val == 1;
}
boolean l = dfs(root.left), r = dfs(root.right);
if (root.val == 2) {
return l || r;
return evaluateTree(root.left) || evaluateTree(root.right);
}
return l && r;
return evaluateTree(root.left) && evaluateTree(root.right);
}
}
```
Expand All @@ -166,14 +156,13 @@ class Solution {
class Solution {
public:
bool evaluateTree(TreeNode* root) {
return dfs(root);
}

bool dfs(TreeNode* root) {
if (!root->left && !root->right) return root->val;
bool l = dfs(root->left), r = dfs(root->right);
if (root->val == 2) return l || r;
return l && r;
if (!root->left) {
return root->val;
}
if (root->val == 2) {
return evaluateTree(root->left) || evaluateTree(root->right);
}
return evaluateTree(root->left) && evaluateTree(root->right);
}
};
```
Expand All @@ -190,18 +179,14 @@ public:
* }
*/
func evaluateTree(root *TreeNode) bool {
var dfs func(*TreeNode) bool
dfs = func(root *TreeNode) bool {
if root.Left == nil && root.Right == nil {
return root.Val == 1
}
l, r := dfs(root.Left), dfs(root.Right)
if root.Val == 2 {
return l || r
}
return l && r
if root.Left == nil {
return root.Val == 1
}
if root.Val == 2 {
return evaluateTree(root.Left) || evaluateTree(root.Right)
} else {
return evaluateTree(root.Left) && evaluateTree(root.Right)
}
return dfs(root)
}
```

Expand All @@ -224,7 +209,7 @@ func evaluateTree(root *TreeNode) bool {

function evaluateTree(root: TreeNode | null): boolean {
const { val, left, right } = root;
if (left == null) {
if (left === null) {
return val === 1;
}
if (val === 2) {
Expand Down Expand Up @@ -257,20 +242,23 @@ function evaluateTree(root: TreeNode | null): boolean {
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>) -> bool {
let root = root.as_ref().unwrap().as_ref().borrow();
if root.left.is_none() {
return root.val == 1;
}
if root.val == 2 {
return Self::dfs(&root.left) || Self::dfs(&root.right);
}
Self::dfs(&root.left) && Self::dfs(&root.right)
}

impl Solution {
pub fn evaluate_tree(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
Self::dfs(&root)
match root {
Some(node) => {
let node = node.borrow();
if node.left.is_none() {
return node.val == 1;
}
if node.val == 2 {
return Self::evaluate_tree(node.left.clone())
|| Self::evaluate_tree(node.right.clone());
}
Self::evaluate_tree(node.left.clone()) && Self::evaluate_tree(node.right.clone())
}
None => false,
}
}
}
```
Expand Down Expand Up @@ -301,112 +289,4 @@ bool evaluateTree(struct TreeNode* root) {

<!-- solution:end -->

<!-- solution:start -->

### 方法二

<!-- tabs:start -->

#### Python3

```python
# 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 evaluateTree(self, root: Optional[TreeNode]) -> bool:
if root.left is None:
return bool(root.val)
l = self.evaluateTree(root.left)
r = self.evaluateTree(root.right)
return l or r if root.val == 2 else l and r
```

#### Java

```java
/**
* 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 boolean evaluateTree(TreeNode root) {
if (root.left == null) {
return root.val == 1;
}
boolean l = evaluateTree(root.left);
boolean r = evaluateTree(root.right);
return root.val == 2 ? l || r : l && r;
}
}
```

#### C++

```cpp
/**
* 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) {}
* };
*/
class Solution {
public:
bool evaluateTree(TreeNode* root) {
if (!root->left) {
return root->val;
}
bool l = evaluateTree(root->left);
bool r = evaluateTree(root->right);
return root->val == 2 ? l or r : l and r;
}
};
```

#### Go

```go
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func evaluateTree(root *TreeNode) bool {
if root.Left == nil {
return root.Val == 1
}
l, r := evaluateTree(root.Left), evaluateTree(root.Right)
if root.Val == 2 {
return l || r
}
return l && r
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->
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