题解 | #对称的二叉树# | Rust

/**
 * #[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{}
    }

    fn dfs(&self, one: Option<Box<TreeNode>>, two: Option<Box<TreeNode>>) -> bool {
        if one.is_none() && two.is_none() {
            return true;
        }
        if one.is_none() || two.is_none() {
            return false;
        }
        return one.as_ref().unwrap().val == two.as_ref().unwrap().val &&
               Solution::dfs(self, one.as_ref().unwrap().left.clone(), two.as_ref().unwrap().right.clone()) &&
               Solution::dfs(self, one.as_ref().unwrap().right.clone(), two.as_ref().unwrap().left.clone());
    }

    /**
    * 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 bool布尔型
    */
    pub fn isSymmetrical(&self, pRoot: Option<Box<TreeNode>>) -> bool {
        return Solution::dfs(self, pRoot.clone(), pRoot.clone());
    }
}