题解 | #二叉树中和为某一值的路径(一)# | 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 root TreeNode类 
        * @param sum int整型 
        * @return bool布尔型
    */
    pub fn hasPathSum(&self, root: Option<Box<TreeNode>>, sum: i32) -> bool {
        if root.is_none() {
            return false;
        }
        if root.as_ref().unwrap().left.is_none() && root.as_ref().unwrap().right.is_none() {
            return sum == root.as_ref().unwrap().val;
        }
        let sum = sum - root.as_ref().unwrap().val;
        return Solution::hasPathSum(self, root.as_ref().unwrap().left.clone(), sum) ||
               Solution::hasPathSum(self, root.as_ref().unwrap().right.clone(), sum);
    }
}