[LeetCode Road] Path Sum - Solution/C++

112. Path Sum

Question:

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.

Source code:

Version 1

Idea:
As the previous questions of binary tree, you must set termination(return) condition when root node is NULL.
Then, you must check a current node of value wheather it is equal to sum when you are at leaf node. (The rule is: go through the left/right node for each degrees, root->val subtracted from sum)

E.g., if you arrival at 7(path: 5->4->11->7), this node is a leaf node (any node whose left and right children are null), a value of sum must be 7.

/**
 * 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 sum) {
        if(!root) return false;
        if(!root->left && !root->right && (root->val == sum)) return true;
        return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
    }
};