notes/OJ notes/pages/Leetcode Path-Sum.md

# Leetcode Path-Sum

#### 2022-07-06 13:45

> ##### Algorithms:
> #algorithm #DFS #recursion 
> ##### Data structures:
> #DS #binary_tree 
> ##### Difficulty:
> #coding_problem #difficulty-easy 
> ##### Additional tags:
> #leetcode 
> ##### Revisions:
> N/A

##### Related topics:
```expander
tag:#DFS
```
 
- [[Leetcode Binary-Tree-Inorder-Traversal]]
- [[Leetcode Binary-Tree-Postorder-Traversal]]
- [[Leetcode Binary-Tree-Preorder-Traversal]]
- [[Leetcode Insert-Into-a-Binary-Search-Tree]]
- [[Leetcode Invert-Binary-Tree]]
- [[Leetcode Lowest-Common-Ancestor-Of-a-Binary-Search-Tree]]
- [[Leetcode Maximum-Depth-Of-Binary-Tree]]
- [[Leetcode Search-In-a-Binary-Tree]]
- [[Leetcode Symmetric-Tree]]
- [[Leetcode Validate-Binary-Search-Tree]]
 

##### Links:
- [Link to problem](https://leetcode.com/problems/path-sum/)
___
### 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.

#### Examples

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/01/18/pathsum1.jpg)

**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:**

![](https://assets.leetcode.com/uploads/2021/01/18/pathsum2.jpg)

**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`

### Thoughts

> [!summary]
> This is a #DFS recursion problem.

There are one thing to consider, return false when the tree is empty.

Simple DFS-like recursion problem.

Base cases:
- node is empty, return false
- node is leaf
	- if the value is sum, return true
	- else return false

Pseudo-code:
- Check for base-cases
- return check(left, sum - root->val) || check(right, sum - root->val)

> [!tip] Why use OR
> By using OR operator, return true when there is at least one  solution that matches.

### Solution

```cpp
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  bool hasPathSum(TreeNode* root, int targetSum) {
    // DFS In-order Recursion
    
    // Base case: node does not exist
    if (!root) {
      return false;
    }
    int val = root->val;
    
    // Base case: reached leaf
    if (!root->left && !root->right) {
      if (targetSum == val)
        return true;
      else 
        return false;
    }
    
    return hasPathSum(root->left, targetSum - val) || hasPathSum(root->right, targetSum - val);
  }
};
```