Path Sum.cpp

/* Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
* Note: A leaf is a node with no children.
*
* Example:
* Given the below binary tree and sum = 22,
*
*       5
*      / \
*     4   8
*    /   / \
*   11  13  4
*  /  \      \
* 7    2      1
*
* return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
*/

/*
 * 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) {}
 * };
 */
 
//Time complexity: O(N). Space complexity: O(N).
class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {
       if(root == NULL){
            return false;
        }else if (root->left == NULL && root->right == NULL && sum - root->val == 0){
            return true;
        }else{
           return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
        }
    }
};