0100 - 0199
0112 - Path Sum (Easy)
Problem Link
https://leetcode.com/problems/path-sum/
Problem Statement
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
Approach 1: DFS
/**
* 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) {
// return false if the root is nullptr
if(!root) return false;
// if it reaches to the end and the val is equal to the sum, return true
if(!root->left && !root->right && root->val == targetSum) return true;
// otherwise, we traverse left node and right node with targetSum `sum - root->val`
return hasPathSum(root->left, targetSum - root->val) || hasPathSum(root->right, targetSum-root->val);
}
};