题解 | #二叉树的直径# | Rust
二叉树的直径
https://www.nowcoder.com/practice/15f977cedc5a4ffa8f03a3433d18650d
/**
* #[derive(PartialEq, Eq, Debug, Clone)]
* pub struct TreeNode {
* pub val: i32,
* pub left: Option<Box<TreeNode>>,
* pub right: Option<Box<TreeNode>>,
* }
*
* impl TreeNode {
* #[inline]
* fn new(val: i32) -> Self {
* TreeNode {
* val: val,
* left: None,
* right: None,
* }
* }
* }
*/
struct Solution{
}
impl Solution {
fn new() -> Self {
Solution{}
}
/**
* 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
*
*
* @param root TreeNode类
* @return int整型
*/
pub fn diameterOfBinaryTree(&self, root: Option<Box<TreeNode>>) -> i32 {
let mut ans: i32 = 0;
Solution::dfs(self, root, &mut ans);
return ans;
}
fn dfs(&self, node: Option<Box<TreeNode>>, ans: &mut i32) -> i32 {
if node.is_none() {
return 0;
}
let mut node = node;
let l = Solution::dfs(self, node.as_mut().unwrap().left.take(), ans);
let r = Solution::dfs(self, node.as_mut().unwrap().right.take(), ans);
*ans = std::cmp::max(*ans, l+r);
return 1 + std::cmp::max(l, r);
}
}
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