vault backup: 2022-07-07 08:06:14

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CS notes/pages/Leetcode Path-Sum.md

@@ -15,23 +15,109 @@
 
 ##### Related topics:
 ```expander
-tag:#<INSERT_TAG_HERE>
+tag:#DFS
 ```
  
  
 
 ##### Links:
-- [Link to problem]()
+- [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 #template_remove_me
+> 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);
+  }
+};
+```