题解 | #二叉树的深度#rust
二叉树的深度
https://www.nowcoder.com/practice/435fb86331474282a3499955f0a41e8b
/** * #[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 pRoot TreeNode类 * @return int整型 */ pub fn TreeDepth(&self, pRoot: Option<Box<TreeNode>>) -> i32 { // write code here // 空间复杂度O(n) 时间复杂度O(n) if pRoot.is_none() { return 0; } let mut left_height = 0; let mut right_height = 0; if let Some(root) = pRoot{ // Box类型,有自动deref功能,所以可以直接写root.left,不需要*root.left left_height = self.TreeDepth(root.left)+1; right_height = self.TreeDepth(root.right)+1; } if left_height > right_height { left_height } else { right_height } } }#rust#