leetcode简单题24 N.111 二叉树的最小深度 rust描述
// [3,9,20,null,null,15,7] 2
// [2,null,3,null,4,null,5,null,6] 5
use std::rc::Rc;
use std::cell::RefCell;
// Definition for a binary tree node.
#[derive(Debug, PartialEq, Eq)]
pub struct TreeNode {
pub val: i32,
pub left: Option<Rc<RefCell<TreeNode>>>,
pub right: Option<Rc<RefCell<TreeNode>>>,
}
impl TreeNode {
#[inline]
pub fn new(val: i32) -> Self {
TreeNode {
val,
left: None,
right: None
}
}
}
// dfs
pub fn min_depth(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
fn min_depth_helper(node: &Option<Rc<RefCell<TreeNode>>>) -> i32 {
match node {
Some(node) => {
let left_depth = min_depth_helper(&node.borrow().left);
let right_depth = min_depth_helper(&node.borrow().right);
if left_depth == 0 || right_depth == 0 {
return left_depth + right_depth + 1;
}
left_depth.min(right_depth) + 1
}
None => 0,
}
}
min_depth_helper(&root)
}
use std::collections::VecDeque;
// bfs
pub fn min_depth2(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
if root.is_none() {
return 0;
}
let mut queue = VecDeque::new();
queue.push_back((root, 1));
while let Some((node, depth)) = queue.pop_front() {
if let Some(node) = node {
let node = node.borrow();
if node.left.is_none() && node.right.is_none() {
return depth;
}
if node.left.is_some() {
queue.push_back((node.left.clone(), depth + 1));
}
if node.right.is_some() {
queue.push_back((node.right.clone(), depth + 1));
}
}
}
0
}
fn main() {}
原文地址:https://blog.csdn.net/BECOMEviolet/article/details/140420935
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