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Jan 15, 2025
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feat: add solutions to lc problems: No.0965,0978 #3954

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80 changes: 49 additions & 31 deletions solution/0900-0999/0965.Univalued Binary Tree/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -56,7 +56,15 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一：DFS

我们记根节点的值为 $x$，然后设计一个函数 $\text{dfs}(\text{root})$，它表示当前节点的值是否等于 $x$，并且它的左右子树也是单值二叉树。

在函数 $\text{dfs}(\text{root})$ 中，如果当前节点为空，那么返回 $\text{true}$，否则，如果当前节点的值等于 $x$，并且它的左右子树也是单值二叉树，那么返回 $\text{true}$，否则返回 $\text{false}$。

在主函数中，我们调用 $\text{dfs}(\text{root})$，并返回结果。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是树中的节点数目。

<!-- tabs:start -->

Expand All @@ -70,12 +78,13 @@ tags:
# self.left = left
# self.right = right
class Solution:
def isUnivalTree(self, root: TreeNode) -> bool:
def dfs(node):
if node is None:
def isUnivalTree(self, root: Optional[TreeNode]) -> bool:
def dfs(root: Optional[TreeNode]) -> bool:
if root is None:
return True
return node.val == root.val and dfs(node.left) and dfs(node.right)
return root.val == x and dfs(root.left) and dfs(root.right)

x = root.val
return dfs(root)
```

Expand All @@ -98,15 +107,18 @@ class Solution:
* }
*/
class Solution {
private int x;

public boolean isUnivalTree(TreeNode root) {
return dfs(root, root.val);
x = root.val;
return dfs(root);
}

private boolean dfs(TreeNode root, int val) {
private boolean dfs(TreeNode root) {
if (root == null) {
return true;
}
return root.val == val && dfs(root.left, val) && dfs(root.right, val);
return root.val == x && dfs(root.left) && dfs(root.right);
}
}
```
Expand All @@ -128,12 +140,14 @@ class Solution {
class Solution {
public:
bool isUnivalTree(TreeNode* root) {
return dfs(root, root->val);
}

bool dfs(TreeNode* root, int val) {
if (!root) return true;
return root->val == val && dfs(root->left, val) && dfs(root->right, val);
int x = root->val;
auto dfs = [&](this auto&& dfs, TreeNode* root) -> bool {
if (!root) {
return true;
}
return root->val == x && dfs(root->left) && dfs(root->right);
};
return dfs(root);
}
};
```
Expand All @@ -150,12 +164,13 @@ public:
* }
*/
func isUnivalTree(root *TreeNode) bool {
x := root.Val
var dfs func(*TreeNode) bool
dfs = func(node *TreeNode) bool {
if node == nil {
dfs = func(root *TreeNode) bool {
if root == nil {
return true
}
return node.Val == root.Val && dfs(node.Left) && dfs(node.Right)
return root.Val == x && dfs(root.Left) && dfs(root.Right)
}
return dfs(root)
}
Expand All @@ -179,14 +194,14 @@ func isUnivalTree(root *TreeNode) bool {
*/

function isUnivalTree(root: TreeNode | null): boolean {
const val = root.val;
const dfs = (root: TreeNode | null) => {
if (root == null) {
const x = root!.val;
const dfs = (root: TreeNode | null): boolean => {
if (!root) {
return true;
}
return root.val === val && dfs(root.left) && dfs(root.right);
return root.val === x && dfs(root.left) && dfs(root.right);
};
return dfs(root.left) && dfs(root.right);
return dfs(root);
}
```

Expand Down Expand Up @@ -214,16 +229,19 @@ function isUnivalTree(root: TreeNode | null): boolean {
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
fn dfs(val: i32, root: &Option<Rc<RefCell<TreeNode>>>) -> bool {
if root.is_none() {
return true;
}
let root = root.as_ref().unwrap().borrow();
root.val == val && Self::dfs(val, &root.left) && Self::dfs(val, &root.right)
}
pub fn is_unival_tree(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
let root = root.as_ref().unwrap().borrow();
Self::dfs(root.val, &root.left) && Self::dfs(root.val, &root.right)
let x = root.as_ref().unwrap().borrow().val;

fn dfs(node: Option<Rc<RefCell<TreeNode>>>, x: i32) -> bool {
if let Some(n) = node {
let n = n.borrow();
n.val == x && dfs(n.left.clone(), x) && dfs(n.right.clone(), x)
} else {
true
}
}

dfs(root, x)
}
}
```
Expand Down
80 changes: 49 additions & 31 deletions solution/0900-0999/0965.Univalued Binary Tree/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -52,7 +52,15 @@ tags:

<!-- solution:start -->

### Solution 1
### Solution 1: DFS

We denote the value of the root node as $x$, and then design a function $\text{dfs}(\text{root})$, which indicates whether the current node's value is equal to $x$ and its left and right subtrees are also univalued binary trees.

In the function $\text{dfs}(\text{root})$, if the current node is null, return $\text{true}$; otherwise, if the current node's value is equal to $x$ and its left and right subtrees are also univalued binary trees, return $\text{true}$; otherwise, return $\text{false}$.

In the main function, we call $\text{dfs}(\text{root})$ and return the result.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the number of nodes in the tree.

<!-- tabs:start -->

Expand All @@ -66,12 +74,13 @@ tags:
# self.left = left
# self.right = right
class Solution:
def isUnivalTree(self, root: TreeNode) -> bool:
def dfs(node):
if node is None:
def isUnivalTree(self, root: Optional[TreeNode]) -> bool:
def dfs(root: Optional[TreeNode]) -> bool:
if root is None:
return True
return node.val == root.val and dfs(node.left) and dfs(node.right)
return root.val == x and dfs(root.left) and dfs(root.right)

x = root.val
return dfs(root)
```

Expand All @@ -94,15 +103,18 @@ class Solution:
* }
*/
class Solution {
private int x;

public boolean isUnivalTree(TreeNode root) {
return dfs(root, root.val);
x = root.val;
return dfs(root);
}

private boolean dfs(TreeNode root, int val) {
private boolean dfs(TreeNode root) {
if (root == null) {
return true;
}
return root.val == val && dfs(root.left, val) && dfs(root.right, val);
return root.val == x && dfs(root.left) && dfs(root.right);
}
}
```
Expand All @@ -124,12 +136,14 @@ class Solution {
class Solution {
public:
bool isUnivalTree(TreeNode* root) {
return dfs(root, root->val);
}

bool dfs(TreeNode* root, int val) {
if (!root) return true;
return root->val == val && dfs(root->left, val) && dfs(root->right, val);
int x = root->val;
auto dfs = [&](this auto&& dfs, TreeNode* root) -> bool {
if (!root) {
return true;
}
return root->val == x && dfs(root->left) && dfs(root->right);
};
return dfs(root);
}
};
```
Expand All @@ -146,12 +160,13 @@ public:
* }
*/
func isUnivalTree(root *TreeNode) bool {
x := root.Val
var dfs func(*TreeNode) bool
dfs = func(node *TreeNode) bool {
if node == nil {
dfs = func(root *TreeNode) bool {
if root == nil {
return true
}
return node.Val == root.Val && dfs(node.Left) && dfs(node.Right)
return root.Val == x && dfs(root.Left) && dfs(root.Right)
}
return dfs(root)
}
Expand All @@ -175,14 +190,14 @@ func isUnivalTree(root *TreeNode) bool {
*/

function isUnivalTree(root: TreeNode | null): boolean {
const val = root.val;
const dfs = (root: TreeNode | null) => {
if (root == null) {
const x = root!.val;
const dfs = (root: TreeNode | null): boolean => {
if (!root) {
return true;
}
return root.val === val && dfs(root.left) && dfs(root.right);
return root.val === x && dfs(root.left) && dfs(root.right);
};
return dfs(root.left) && dfs(root.right);
return dfs(root);
}
```

Expand Down Expand Up @@ -210,16 +225,19 @@ function isUnivalTree(root: TreeNode | null): boolean {
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
fn dfs(val: i32, root: &Option<Rc<RefCell<TreeNode>>>) -> bool {
if root.is_none() {
return true;
}
let root = root.as_ref().unwrap().borrow();
root.val == val && Self::dfs(val, &root.left) && Self::dfs(val, &root.right)
}
pub fn is_unival_tree(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
let root = root.as_ref().unwrap().borrow();
Self::dfs(root.val, &root.left) && Self::dfs(root.val, &root.right)
let x = root.as_ref().unwrap().borrow().val;

fn dfs(node: Option<Rc<RefCell<TreeNode>>>, x: i32) -> bool {
if let Some(n) = node {
let n = n.borrow();
n.val == x && dfs(n.left.clone(), x) && dfs(n.right.clone(), x)
} else {
true
}
}

dfs(root, x)
}
}
```
Expand Down
16 changes: 9 additions & 7 deletions solution/0900-0999/0965.Univalued Binary Tree/Solution.cpp
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -12,11 +12,13 @@
class Solution {
public:
bool isUnivalTree(TreeNode* root) {
return dfs(root, root->val);
int x = root->val;
auto dfs = [&](this auto&& dfs, TreeNode* root) -> bool {
if (!root) {
return true;
}
return root->val == x && dfs(root->left) && dfs(root->right);
};
return dfs(root);
}

bool dfs(TreeNode* root, int val) {
if (!root) return true;
return root->val == val && dfs(root->left, val) && dfs(root->right, val);
}
};
};
9 changes: 5 additions & 4 deletions solution/0900-0999/0965.Univalued Binary Tree/Solution.go
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Original file line number Diff line number Diff line change
Expand Up @@ -7,12 +7,13 @@
* }
*/
func isUnivalTree(root *TreeNode) bool {
x := root.Val
var dfs func(*TreeNode) bool
dfs = func(node *TreeNode) bool {
if node == nil {
dfs = func(root *TreeNode) bool {
if root == nil {
return true
}
return node.Val == root.Val && dfs(node.Left) && dfs(node.Right)
return root.Val == x && dfs(root.Left) && dfs(root.Right)
}
return dfs(root)
}
}
11 changes: 7 additions & 4 deletions solution/0900-0999/0965.Univalued Binary Tree/Solution.java
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Original file line number Diff line number Diff line change
Expand Up @@ -14,14 +14,17 @@
* }
*/
class Solution {
private int x;

public boolean isUnivalTree(TreeNode root) {
return dfs(root, root.val);
x = root.val;
return dfs(root);
}

private boolean dfs(TreeNode root, int val) {
private boolean dfs(TreeNode root) {
if (root == null) {
return true;
}
return root.val == val && dfs(root.left, val) && dfs(root.right, val);
return root.val == x && dfs(root.left) && dfs(root.right);
}
}
}
9 changes: 5 additions & 4 deletions solution/0900-0999/0965.Univalued Binary Tree/Solution.py
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Original file line number Diff line number Diff line change
Expand Up @@ -5,10 +5,11 @@
# self.left = left
# self.right = right
class Solution:
def isUnivalTree(self, root: TreeNode) -> bool:
def dfs(node):
if node is None:
def isUnivalTree(self, root: Optional[TreeNode]) -> bool:
def dfs(root: Optional[TreeNode]) -> bool:
if root is None:
return True
return node.val == root.val and dfs(node.left) and dfs(node.right)
return root.val == x and dfs(root.left) and dfs(root.right)

x = root.val
return dfs(root)
21 changes: 12 additions & 9 deletions solution/0900-0999/0965.Univalued Binary Tree/Solution.rs
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Original file line number Diff line number Diff line change
Expand Up @@ -19,15 +19,18 @@
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
fn dfs(val: i32, root: &Option<Rc<RefCell<TreeNode>>>) -> bool {
if root.is_none() {
return true;
}
let root = root.as_ref().unwrap().borrow();
root.val == val && Self::dfs(val, &root.left) && Self::dfs(val, &root.right)
}
pub fn is_unival_tree(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
let root = root.as_ref().unwrap().borrow();
Self::dfs(root.val, &root.left) && Self::dfs(root.val, &root.right)
let x = root.as_ref().unwrap().borrow().val;

fn dfs(node: Option<Rc<RefCell<TreeNode>>>, x: i32) -> bool {
if let Some(n) = node {
let n = n.borrow();
n.val == x && dfs(n.left.clone(), x) && dfs(n.right.clone(), x)
} else {
true
}
}

dfs(root, x)
}
}
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