Struct leetcode_test_utils::tree::diagnosis::T

pub struct T(pub TreeHandle);

Zero cost wrapper for Option<Rc<RefCell<TreeNode>>>, also for bypassing the orphan rule. There are many useful methods for operating on the binary tree as well.

Implementations

impl T

pub fn height(&self) -> usize

Get the height for the binary tree.

Returns the height of the binary tree. The empty tree has height 0.

Examples

use leetcode_test_utils::btree;
use leetcode_test_utils::tree::T;

let tree1 = btree!(1, 2, 3);
assert_eq!(T(tree1).height(), 2);
assert_eq!(T(None).height(), 0);

pub fn pre_order(&self) -> Vec<i32>

Returns the pre-order of the binary tree.

pub fn in_order(&self) -> Vec<i32>

Returns the in-order of the binary tree.

pub fn post_order(&self) -> Vec<i32>

Returns the post-order of the binary tree.

pub fn level_order(&self) -> Vec<i32>

Returns the level order of the binary tree.

pub fn re_owned(&mut self)

Launder the binary tree.

Replace the current binary tree with a new representation, in which the structure and values is preserved respectively, but every reachable Rc will only have 1 strong count.

This is helpful if you do not want the value in your tree changed through Rc<RefCell<TreeNode>> elsewhere.

Examples

use leetcode_test_utils::btree;
use leetcode_test_utils::tree::T;
use std::rc::Rc;

let tree = T(btree!(3));
let evil = Rc::clone(tree.0.as_ref().unwrap());
// the action below changes the value handled in `tree`, which may be unexpected
evil.borrow_mut().val = 42;
assert_ne!(tree.0.unwrap().borrow().val, 3);

pub fn detach(&self) -> Self

Get the mirror tree.

Returns a binary tree sharing the same structure and values handled by self except that every reachable Rc will only have 1 strong count.

This is helpful if you want to get the tree structure without worrying about the values be soon changed by code elsewhere.

Examples

use leetcode_test_utils::btree;
use leetcode_test_utils::tree::T;
use std::rc::Rc;

let tree = T(btree!(3));
let cannot_invade = tree.detach();
cannot_invade.0.as_ref().unwrap().borrow_mut().val = 42;
assert_eq!(tree.0.unwrap().borrow().val, 3);

pub fn is_balanced(&self) -> bool

Test if the binary tree is balanced.

Example

use leetcode_test_utils::btree;
use leetcode_test_utils::tree::T;

let tree1 = T(btree!(4, 2));
assert!(tree1.is_balanced());

let tree2 = T(btree!(4, 2, null, 1));
assert!(!tree2.is_balanced())

pub fn is_binary_search_tree(&self) -> bool

Test if the binary tree is a BST(binary search tree).

Examples

use leetcode_test_utils::btree;
use leetcode_test_utils::tree::T;

let tree = T(btree!(5, 2, 9, 1));
assert!(tree.is_binary_search_tree());

pub fn to_leetcode_raw(&self) -> String

Returns the leetcode representation of the handled binary tree.

Examples

use leetcode_test_utils::tree::T;
use leetcode_test_utils::btree;

let tree = T(btree!(2, 4, null, 9));
assert_eq!(tree.to_leetcode_raw(), "[2,4,null,9]");

pub fn to_leetcode(&self) -> Vec<Option<i32>>

Returns the parsed leetcode representation of the handled binary tree.

Examples

use leetcode_test_utils::tree::T;
use leetcode_test_utils::btree;

let tree = T(btree!(2, 4, null, 9));
assert_eq!(tree.to_leetcode(), vec![Some(2), Some(4), None, Some(9)]);

pub fn len(&self) -> usize

Returns the len of the binary tree.

Examples

use leetcode_test_utils::tree::T;
use leetcode_test_utils::btree;

let tree = T(btree!(2, 4, null, 9));
assert_eq!(tree.len(), 3);

Trait Implementations

impl Clone for T

fn clone(&self) -> T

Returns a copy of the value.

fn clone_from(&mut self, source: &Self)

Performs copy-assignment from source.

impl Debug for T

fn fmt(&self, f: &mut Formatter<'_>) -> Result

Formats the value using the given formatter.

impl Display for T

fn fmt(&self, f: &mut Formatter<'_>) -> Result

Formats the value using the given formatter.

impl PartialEq<T> for T

fn eq(&self, other: &Self) -> bool

This method tests for self and other values to be equal, and is used by ==.

fn ne(&self, other: &Rhs) -> bool

This method tests for !=.