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feat: add rust solution to lc problem: No.0894 (doocs#1274)
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‎solution/0800-0899/0894.All Possible Full Binary Trees/README.md

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};
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```
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### **Rust**
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```rust
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// Definition for a binary tree node.
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// #[derive(Debug, PartialEq, Eq)]
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// pub struct TreeNode {
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// pub val: i32,
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// pub left: Option<Rc<RefCell<TreeNode>>>,
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// pub right: Option<Rc<RefCell<TreeNode>>>,
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// }
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//
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// impl TreeNode {
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// #[inline]
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// pub fn new(val: i32) -> Self {
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// TreeNode {
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// val,
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// left: None,
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// right: None
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// }
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// }
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// }
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impl TreeNode {
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pub fn new_with_node(left: Option<Rc<RefCell<TreeNode>>>, right: Option<Rc<RefCell<TreeNode>>>) -> Self {
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Self {
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val: 0,
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left,
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right,
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}
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}
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}
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use std::rc::Rc;
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use std::cell::RefCell;
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impl Solution {
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#[allow(dead_code)]
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pub fn all_possible_fbt(n: i32) -> Vec<Option<Rc<RefCell<TreeNode>>>> {
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let mut record_vec = vec![vec![]; n as usize + 1];
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Self::dfs(n, &mut record_vec)
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}
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#[allow(dead_code)]
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fn dfs(n: i32, record_vec: &mut Vec<Vec<Option<Rc<RefCell<TreeNode>>>>>) -> Vec<Option<Rc<RefCell<TreeNode>>>> {
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if record_vec[n as usize].len() != 0 {
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return record_vec[n as usize].clone();
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}
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if n == 1 {
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// Just directly return a single node
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return vec![Some(Rc::new(RefCell::new(TreeNode::new(0))))];
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}
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// Otherwise, need to construct return vector
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let mut ret_vec = Vec::new();
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// Enumerate the node number for left subtree from 0 -> n - 1
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for i in 0..n - 1 {
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// The number of right subtree node
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let j = n - i - 1;
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for left in Self::dfs(i, record_vec) {
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for right in Self::dfs(j, record_vec) {
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// Construct the ret vector
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ret_vec.push(Some(Rc::new(RefCell::new(TreeNode::new_with_node(left.clone(), right.clone())))));
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}
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}
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}
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record_vec[n as usize] = ret_vec;
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record_vec[n as usize].clone()
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}
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}
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```
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### **Go**
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```go

‎solution/0800-0899/0894.All Possible Full Binary Trees/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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// Definition for a binary tree node.
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// #[derive(Debug, PartialEq, Eq)]
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// pub struct TreeNode {
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// pub val: i32,
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// pub left: Option<Rc<RefCell<TreeNode>>>,
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// pub right: Option<Rc<RefCell<TreeNode>>>,
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// }
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//
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// impl TreeNode {
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// #[inline]
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// pub fn new(val: i32) -> Self {
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// TreeNode {
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// val,
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// left: None,
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// right: None
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// }
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// }
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// }
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impl TreeNode {
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pub fn new_with_node(left: Option<Rc<RefCell<TreeNode>>>, right: Option<Rc<RefCell<TreeNode>>>) -> Self {
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Self {
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val: 0,
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left,
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right,
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}
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}
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}
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use std::rc::Rc;
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use std::cell::RefCell;
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impl Solution {
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#[allow(dead_code)]
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pub fn all_possible_fbt(n: i32) -> Vec<Option<Rc<RefCell<TreeNode>>>> {
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let mut record_vec = vec![vec![]; n as usize + 1];
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Self::dfs(n, &mut record_vec)
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}
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#[allow(dead_code)]
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fn dfs(n: i32, record_vec: &mut Vec<Vec<Option<Rc<RefCell<TreeNode>>>>>) -> Vec<Option<Rc<RefCell<TreeNode>>>> {
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if record_vec[n as usize].len() != 0 {
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return record_vec[n as usize].clone();
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}
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if n == 1 {
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// Just directly return a single node
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return vec![Some(Rc::new(RefCell::new(TreeNode::new(0))))];
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}
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// Otherwise, need to construct return vector
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let mut ret_vec = Vec::new();
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// Enumerate the node number for left subtree from 0 -> n - 1
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for i in 0..n - 1 {
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// The number of right subtree node
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let j = n - i - 1;
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for left in Self::dfs(i, record_vec) {
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for right in Self::dfs(j, record_vec) {
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// Construct the ret vector
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ret_vec.push(Some(Rc::new(RefCell::new(TreeNode::new_with_node(left.clone(), right.clone())))));
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}
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}
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}
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record_vec[n as usize] = ret_vec;
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record_vec[n as usize].clone()
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}
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}
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```
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### **Go**
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```go

‎solution/0800-0899/0894.All Possible Full Binary Trees/Solution.rs

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// Definition for a binary tree node.
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// #[derive(Debug, PartialEq, Eq)]
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// pub struct TreeNode {
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// pub val: i32,
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// pub left: Option<Rc<RefCell<TreeNode>>>,
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// pub right: Option<Rc<RefCell<TreeNode>>>,
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// }
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//
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// impl TreeNode {
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// #[inline]
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// pub fn new(val: i32) -> Self {
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// TreeNode {
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// val,
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// left: None,
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// right: None
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// }
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// }
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// }
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impl TreeNode {
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pub fn new_with_node(left: Option<Rc<RefCell<TreeNode>>>, right: Option<Rc<RefCell<TreeNode>>>) -> Self {
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Self {
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val: 0,
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left,
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right,
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}
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}
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}
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use std::rc::Rc;
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use std::cell::RefCell;
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impl Solution {
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#[allow(dead_code)]
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pub fn all_possible_fbt(n: i32) -> Vec<Option<Rc<RefCell<TreeNode>>>> {
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let mut record_vec = vec![vec![]; n as usize + 1];
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Self::dfs(n, &mut record_vec)
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}
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#[allow(dead_code)]
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fn dfs(n: i32, record_vec: &mut Vec<Vec<Option<Rc<RefCell<TreeNode>>>>>) -> Vec<Option<Rc<RefCell<TreeNode>>>> {
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if record_vec[n as usize].len() != 0 {
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return record_vec[n as usize].clone();
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}
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if n == 1 {
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// Just directly return a single node
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return vec![Some(Rc::new(RefCell::new(TreeNode::new(0))))];
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}
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// Otherwise, need to construct return vector
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let mut ret_vec = Vec::new();
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// Enumerate the node number for left subtree from 0 -> n - 1
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for i in 0..n - 1 {
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// The number of right subtree node
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let j = n - i - 1;
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for left in Self::dfs(i, record_vec) {
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for right in Self::dfs(j, record_vec) {
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// Construct the ret vector
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ret_vec.push(Some(Rc::new(RefCell::new(TreeNode::new_with_node(left.clone(), right.clone())))));
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}
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}
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}
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record_vec[n as usize] = ret_vec;
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record_vec[n as usize].clone()
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}
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}

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