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); } }