题解 | #合并二叉树# | 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{}
    }

    /**
    * 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 t1 TreeNode类 
        * @param t2 TreeNode类 
        * @return TreeNode类
    */
    pub fn mergeTrees(&self, t1: Option<Box<TreeNode>>, t2: Option<Box<TreeNode>>) -> Option<Box<TreeNode>> {
        if t1.is_none() {
            return t2;
        }
        if t2.is_none() {
            return t1;
        }
        let mut t1 = t1;
        let mut t2 = t2;
        t1.as_mut().unwrap().val += t2.as_mut().unwrap().val;
        t1.as_mut().unwrap().left = Solution::mergeTrees(self, t1.as_mut().unwrap().left.take(), t2.as_mut().unwrap().left.take());
        t1.as_mut().unwrap().right = Solution::mergeTrees(self, t1.as_mut().unwrap().right.take(), t2.as_mut().unwrap().right.take());
        return t1;
    }
}