112. Path Sum

Description of the Problem

Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.

A leaf is a node with no children.

Example 1:

Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
Output: true
Explanation: The root-to-leaf path with the target sum is shown.

Example 2:

Input: root = [1,2,3], targetSum = 5
Output: false
Explanation: There two root-to-leaf paths in the tree:
(1 --> 2): The sum is 3.
(1 --> 3): The sum is 4.
There is no root-to-leaf path with sum = 5.

Example 3:

Input: root = [], targetSum = 0
Output: false
Explanation: Since the tree is empty, there are no root-to-leaf paths.

Constraints:

The number of nodes in the tree is in the range [0, 5000].
1000 <= Node.val <= 1000
1000 <= targetSum <= 1000

Solution

Code (Rust)

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
// 
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
    pub fn has_path_sum(root: Option<Rc<RefCell<TreeNode>>>, target_sum: i32) 
        -> bool {
        match (root) {
            None => false,
            Some(root) => {
                let root = root.borrow();
                if root.left.is_none() 
                    && root.right.is_none() 
                    && root.val == target_sum {
                    true
                }else {
                    Solution::has_path_sum( 
                        root.left.clone(), target_sum - root.val
                    ) 
                    || Solution::has_path_sum( 
                        root.right.clone(), target_sum - root.val
                    )
                }
            }
        }
    }
}

Complexity

n is the number of nodes in the tree;
h is the height of the tree

Time complexity:

\( T(n) = O(n) \)

Auxiliary Space:

\( S(n) = O(h) \)

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