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Commit dac50d6

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feat: add rust solution to lc problem: No.0222
No.0222.Count Complete Tree Nodes
1 parent 7e5479a commit dac50d6

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6 files changed

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-5
lines changed

6 files changed

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‎index.html

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<script src="//cdn.jsdelivr.net/npm/docsify-darklight-theme@latest/dist/index.min.js"></script>
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</body>
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</html>
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</html>

‎solution/0200-0299/0222.Count Complete Tree Nodes/README.md

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@@ -270,6 +270,38 @@ public class Solution {
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}
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```
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### **Rust**
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```rust
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use std::cell::RefCell;
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use std::rc::Rc;
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impl Solution {
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pub fn count_nodes(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
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if let Some(node) = root {
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let node = node.borrow();
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let left = Self::depth(&node.left);
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let right = Self::depth(&node.right);
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if left == right {
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Self::count_nodes(node.right.clone()) + (1 << left)
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} else {
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Self::count_nodes(node.left.clone()) + (1 << right)
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}
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} else {
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0
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}
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}
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fn depth(root: &Option<Rc<RefCell<TreeNode>>>) -> i32 {
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if let Some(node) = root {
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Self::depth(&node.borrow().left) + 1
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} else {
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0
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}
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}
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}
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```
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### **...**
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```

‎solution/0200-0299/0222.Count Complete Tree Nodes/README_EN.md

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}
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```
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### **Rust**
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```rust
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use std::cell::RefCell;
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use std::rc::Rc;
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impl Solution {
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pub fn count_nodes(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
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if let Some(node) = root {
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let node = node.borrow();
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let left = Self::depth(&node.left);
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let right = Self::depth(&node.right);
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if left == right {
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Self::count_nodes(node.right.clone()) + (1 << left)
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} else {
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Self::count_nodes(node.left.clone()) + (1 << right)
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}
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} else {
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0
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}
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}
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fn depth(root: &Option<Rc<RefCell<TreeNode>>>) -> i32 {
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if let Some(node) = root {
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Self::depth(&node.borrow().left) + 1
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} else {
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0
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}
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}
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}
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```
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### **...**
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```
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use std::cell::RefCell;
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use std::rc::Rc;
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impl Solution {
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pub fn count_nodes(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
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if let Some(node) = root {
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let node = node.borrow();
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let left = Self::depth(&node.left);
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let right = Self::depth(&node.right);
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if left == right {
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Self::count_nodes(node.right.clone()) + (1 << left)
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} else {
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Self::count_nodes(node.left.clone()) + (1 << right)
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}
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} else {
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0
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}
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}
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fn depth(root: &Option<Rc<RefCell<TreeNode>>>) -> i32 {
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if let Some(node) = root {
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Self::depth(&node.borrow().left) + 1
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} else {
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0
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}
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}
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}

‎solution/2000-2099/2000.Reverse Prefix of Word/README.md

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<pre><strong>输入:</strong>word = "<em><strong>abcd</strong></em>efd", ch = "d"
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<strong>输出:</strong>"<em><strong>dcba</strong></em>efd"
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<strong>解释:</strong>"d" 第一次出现在下标 3 。
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<strong>解释:</strong>"d" 第一次出现在下标 3 。
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反转从下标 0 到下标 3(含下标 3)的这段字符,结果字符串是 "dcbaefd" 。
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</pre>
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### **Rust**
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```rs
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```rust
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impl Solution {
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pub fn reverse_prefix(word: String, ch: char) -> String {
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match word.find(ch) {

‎solution/2000-2099/2000.Reverse Prefix of Word/README_EN.md

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<pre>
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<strong>Input:</strong> word = &quot;<u>abcd</u>efd&quot;, ch = &quot;d&quot;
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<strong>Output:</strong> &quot;<u>dcba</u>efd&quot;
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<strong>Explanation:</strong>&nbsp;The first occurrence of &quot;d&quot; is at index 3.
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<strong>Explanation:</strong>&nbsp;The first occurrence of &quot;d&quot; is at index 3.
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Reverse the part of word from 0 to 3 (inclusive), the resulting string is &quot;dcbaefd&quot;.
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</pre>
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### **Rust**
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```rs
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```rust
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impl Solution {
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pub fn reverse_prefix(word: String, ch: char) -> String {
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match word.find(ch) {

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