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Commit 9c32528

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Merge pull request youngyangyang04#1330 from 3Xpl0it3r/master
添加(102. 二叉树的层序遍历 I) Rust 版本
2 parents dab44f5 + 9611896 commit 9c32528

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‎problems/0102.二叉树的层序遍历.md

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Original file line numberDiff line numberDiff line change
@@ -319,6 +319,36 @@ func levelOrder(_ root: TreeNode?) -> [[Int]] {
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}
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```
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Rust:
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```rust
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pub fn level_order(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {
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let mut ans = Vec::new();
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let mut stack = Vec::new();
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if root.is_none(){
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return ans;
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}
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stack.push(root.unwrap());
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while stack.is_empty()!= true{
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let num = stack.len();
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let mut level = Vec::new();
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for _i in 0..num{
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let tmp = stack.remove(0);
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level.push(tmp.borrow_mut().val);
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if tmp.borrow_mut().left.is_some(){
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stack.push(tmp.borrow_mut().left.take().unwrap());
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}
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if tmp.borrow_mut().right.is_some(){
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stack.push(tmp.borrow_mut().right.take().unwrap());
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}
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}
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ans.push(level);
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}
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ans
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}
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```
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**此时我们就掌握了二叉树的层序遍历了,那么如下九道力扣上的题目,只需要修改模板的两三行代码(不能再多了),便可打倒!**
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@@ -548,6 +578,35 @@ func levelOrderBottom(_ root: TreeNode?) -> [[Int]] {
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}
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```
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Rust:
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```rust
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pub fn level_order(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {
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let mut ans = Vec::new();
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let mut stack = Vec::new();
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if root.is_none(){
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return ans;
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}
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stack.push(root.unwrap());
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while stack.is_empty()!= true{
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let num = stack.len();
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let mut level = Vec::new();
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for _i in 0..num{
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let tmp = stack.remove(0);
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level.push(tmp.borrow_mut().val);
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if tmp.borrow_mut().left.is_some(){
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stack.push(tmp.borrow_mut().left.take().unwrap());
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}
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if tmp.borrow_mut().right.is_some(){
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stack.push(tmp.borrow_mut().right.take().unwrap());
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}
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}
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ans.push(level);
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}
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ans
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}
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```
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# 199.二叉树的右视图
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[力扣题目链接](https://leetcode-cn.com/problems/binary-tree-right-side-view/)

‎problems/0104.二叉树的最大深度.md

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Original file line numberDiff line numberDiff line change
@@ -192,6 +192,33 @@ public:
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};
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```
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rust:
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```rust
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impl Solution {
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pub fn max_depth(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
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if root.is_none(){
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return 0;
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}
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let mut max_depth: i32 = 0;
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let mut stack = vec![root.unwrap()];
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while !stack.is_empty() {
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let num = stack.len();
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for _i in 0..num{
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let top = stack.remove(0);
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if top.borrow_mut().left.is_some(){
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stack.push(top.borrow_mut().left.take().unwrap());
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}
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if top.borrow_mut().right.is_some(){
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stack.push(top.borrow_mut().right.take().unwrap());
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}
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}
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max_depth+=1;
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}
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max_depth
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}
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```
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那么我们可以顺便解决一下n叉树的最大深度问题
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# 559.n叉树的最大深度

‎problems/0111.二叉树的最小深度.md

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Original file line numberDiff line numberDiff line change
@@ -488,5 +488,69 @@ func minDepth(_ root: TreeNode?) -> Int {
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}
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```
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rust:
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```rust
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impl Solution {
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pub fn min_depth(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
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return Solution::bfs(root)
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}
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// 递归
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pub fn dfs(node: Option<Rc<RefCell<TreeNode>>>) -> i32{
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if node.is_none(){
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return 0;
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}
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let parent = node.unwrap();
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let left_child = parent.borrow_mut().left.take();
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let right_child = parent.borrow_mut().right.take();
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if left_child.is_none() && right_child.is_none(){
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return 1;
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}
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let mut min_depth = i32::MAX;
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if left_child.is_some(){
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let left_depth = Solution::dfs(left_child);
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if left_depth <= min_depth{
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min_depth = left_depth
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}
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}
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if right_child.is_some(){
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let right_depth = Solution::dfs(right_child);
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if right_depth <= min_depth{
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min_depth = right_depth
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}
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}
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min_depth + 1
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}
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// 迭代
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pub fn bfs(node: Option<Rc<RefCell<TreeNode>>>) -> i32{
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let mut min_depth = 0;
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if node.is_none(){
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return min_depth
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}
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let mut stack = vec![node.unwrap()];
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while !stack.is_empty(){
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min_depth += 1;
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let num = stack.len();
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for _i in 0..num{
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let top = stack.remove(0);
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let left_child = top.borrow_mut().left.take();
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let right_child = top.borrow_mut().right.take();
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if left_child.is_none() && right_child.is_none(){
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return min_depth;
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}
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if left_child.is_some(){
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stack.push(left_child.unwrap());
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}
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if right_child.is_some(){
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stack.push(right_child.unwrap());
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}
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}
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}
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min_depth
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}
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```
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-----------------------
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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